C語言編程題庫與解析2

經(jīng)典的數(shù)學(xué)問題:1-1、三色球問題:若一個(gè)口袋中放有12個(gè)球,其中有3個(gè)紅色的,3個(gè)白色的,6個(gè)黑色的,從中任取8個(gè)球,問共有多少種不同顏色的搭配?#include<stdio.h>intmain()(inti,kj=O;for(i=0;i<=3;i++)(for(k=0;k<=3;k++)(if(i+k>=2)j++;)}printf("共有％d種方案、n”J);return0;)L2：12個(gè)球中摸出3個(gè)球都是紅色的概率是多少?#include<stdio.h>intmain。|floatiJ,k,sum=0;for(i=0;i<=3;i++)for(j=0;j<=3;j++)for(k=0;k<=6;k++)if(i+k+j==3)sum++;printf(''為％.2f\n'',1/sum);return0;90.10ressanykeytocontinue2、雞兔同籠問題:共有98個(gè)頭,386只腳,編程求雞、兔各多少只?#include<stdio.h>intmain()(intx,y;for(x=1;x<=386;x++)(if(2*x+4*(98-x)==386)(x=x;y=98-x;break;}}printf("雞有％d只,兔有％d只。”,x,y);return0;)3、百雞問題:雞翁ー,值錢五;母雞ー,值錢三;雞雛三,值錢ー。百錢買百雞,問雞翁,母雞,雞雛各幾何?#include<stdio.h>intmain(){inta=O,b,c;while(a<=20)//雞翁個(gè)數(shù)不能超過20個(gè){for(b=0;bv=33;b++)/鴻母個(gè)數(shù)不能超過100/3個(gè)for(c=0;cv=300;c=c+3)〃雞雛個(gè)數(shù)不超過300只,for循環(huán)下的循環(huán)(if(a+b+c==100&&a*5+b*3+c/3==100)//三種雞總共100只價(jià)錢共100printf("雞翁:%d\n雞母:％d\n雞雛:％d\n\n\n”,a,b,c);}a++;〃與前面while循環(huán)聯(lián)系}return0;}1、教材Pill的5.5；#include<stdio.h>voidmain(){doublex,y;printf("請輸入x:つ;scanf("%1f",&x);if(x<l)y=x;elseif(x<10)y=2*x-l;elsey=3*xTl;printf("f(%.2f)=%.2f\n”,x,y);}博輸入X：1f<1.00>=1.00Pressanykeytocontinue2.分析一下程序段輸入的結(jié)果(一定要先紙上,后機(jī)器驗(yàn)證)#include<stdio.h>voidmain(){intx=l,y=-10;if(x<++y)if(y—<0)x+=y;elsex+=3*y;elsex=3*y;printf(*%d\n*,x);)首先判斷x<++y,此時(shí)y=-9,而又不滿足if條件,所以就進(jìn)入else,x=3*y;自然,x=-27.-273、輸入若干整數(shù)給x,使用if語句輸出x的絕對(duì)值,當(dāng)輸入〇時(shí)結(jié)束;??includeくstdio.h>intmain(){intx;printf("請輸入ー個(gè)數(shù)、n");scanf(*%d*,&x);if(x==0)printf("%d的絕對(duì)值是%d\n",x,x);for(;x!=0;)if(x>=0)printf("%d的絕對(duì)值是%d\n”,x,x);elseprintf("%d的絕對(duì)值是%d\n”,x,-x);scanf(*%d*,&x);)return0;MyA一棟8的絕對(duì)值是8卜8-8的絕對(duì)值是80Pressanykeytocontinue4、將輸入字符的大寫英文字母轉(zhuǎn)換成小寫字母輸出,小寫字母原樣輸出#includeくstdio.h>intmain()(chara;printf("請輸入ー組英文字母,并以#結(jié)束:\n")；while((a=getchar())!='#'){if(a<=122,a>=97)putchar(a);elseputchar(a+32);}printf("\n");return0;}5、課件例題6.1#include<stdio.h>intmain(){intx,sum=O;scanf("%d",&x);for(;x;scanf&x))if(x>0)sum=sum+x;printf("正數(shù)和為:%d\n",sum);return0;世數(shù)和為:21Pressanykeytocontinue6、根據(jù)代碼寫出代碼所完成的功能#include<stdio.h>voidmain(){doublex,y;printf("請輸入x：");scanf&x);if(x<l)y=x+l;elseif(x<2)y=x+2;elsey=x+3;printf(*f(%.2f)=%.2f\n*,x,y);)此程序的功能計(jì)算ー個(gè)分段函數(shù)。y=x+lx<ly=x+2x<2y=x+3x>=25、本周本次課程小結(jié)本周的作業(yè)還算簡單。我覺得本人在C上的弱點(diǎn)就是利用程序做數(shù)學(xué)問題。我想我自己對(duì)編程的練習(xí)還不夠深入,練得還不夠多。我認(rèn)為我接下的階段,要把以前做過的作業(yè)都復(fù)習(xí)一遍,總結(jié)出ー些規(guī)律和方法,這樣對(duì)以后的學(xué)習(xí)有幫助,對(duì)期中考試也有幫助。1、鍵盤輸入0-6之間的整數(shù),分別代表每周的星期日、星期ー、…星期六,請編寫ー個(gè)程序,輸入0-6之間的ー個(gè)整數(shù),輸出相應(yīng)的星期。（最好有個(gè)提示）#include<stdio.h>intmain（）Iintday;printf（"請輸入星期的數(shù)字編號(hào)（0-6）:"）;scanf（*%d*,&day）;switch（day）{case0:printf（"對(duì)應(yīng)的星期為星期日\n"）;break;printf（"對(duì)應(yīng)的星期為星期ー、n"）;break;printf（"對(duì)應(yīng)的星期為星期二、n"）;break;printf（"對(duì)應(yīng)的星期為星期三\n"）;break;printf（"對(duì)應(yīng)的星期為星期四ゝn"）;break;printf（"對(duì)應(yīng)的星期為星期五'n"）;break;printf（"對(duì)應(yīng)的星期為星期六、n"）;break;printf（"輸入有誤,請輸入〇ー6之間的數(shù)字［'n"）;）return0;}2、問題的提出:編寫程序,實(shí)現(xiàn)如下功能:輸入ー個(gè)實(shí)數(shù)后,屏幕上顯示如下菜單：輸出相反數(shù)輸出平方數(shù)輸出平方根退出輸入1,輸出該數(shù)的相反數(shù);輸入2,輸出該數(shù)的的平方數(shù);輸入3,輸出該數(shù)的的平方根;輸入4退出此程序。輸入む4之外的數(shù),顯示“請輸入「4之間的數(shù)字”。ttinclude<stdio.h>??include<math.h>

intmainOintc;floatm;printfC 歡迎使用本系統(tǒng)!\n(說明:本系統(tǒng)的功能是輸出相反數(shù)輸出平方數(shù)輸出平方根)'ホ)；printf("請輸入任意ー個(gè)整型數(shù):\n");scanf&m);pnntf( 101t*3Me**jioi6*3ioi￡*\n);printf(*b輸出相反數(shù)、n");printf「2,輸出平方數(shù)、rT);printf(*3,輸出開方根'n");printfC4.退出'n");whiled)|scanfC%d*?&c);switch(c)(printf("輸出相反數(shù):%f\n*?-m);break;printf("輸出平方數(shù):%f\n”,m*m);break;printf("輸出平方根:％f\n",sqrt(m));break;break;default:printf("請輸入1?4之間的數(shù)字!\n");}if(c=4)break;)printf("****************************\n 謝謝使用本系統(tǒng),再見?、n");return0;}輸出平方數(shù)輸出平方根,1234)數(shù)數(shù)根反方方鞏開出出出出退輸出平方數(shù)輸出平方根,1234)數(shù)數(shù)根反方方鞏開出出出出退輸出相反數(shù):-5.000000爐出平方數(shù):

キ出平方根:25.0000002.236068爐出平方數(shù):

キ出平方根:25.0000002.236068謝謝使用本系統(tǒng)，再見?Pressanykeytocontinue3^假設(shè)自動(dòng)售貨機(jī)出售4種商品,薯片(crisps)、爆米花(popcorn)、巧克カ(chocolate)和可樂(cola),售價(jià)分別是每份3.0^2.5、4.0和3.5元。在屏幕上顯示以下菜單,用戶可以連續(xù)查詢商品的價(jià)格,當(dāng)查詢次數(shù)超過5次時(shí),自動(dòng)退出查詢;不到5次時(shí),用戶可以選擇退出。如果假設(shè)買入多種商品,希望在結(jié)算的時(shí)候,有提示,最好有“謝謝光顧”等。#include<stdio.h>^include<math.h>intmain()(inti=0,c,d,e,f,g;while(1)(printf(*****5101s101t*5101t3101t**s1ale;printfCl,價(jià)格査詢系統(tǒng) 2.自動(dòng)售賣系統(tǒng)ゝn");printf("請選擇你要進(jìn)行的操作:.)；scanf("%d",&c);if(c=l)(printf(*\n\n歡迎使用價(jià)格査詢系統(tǒng)\n");printfCl,JPt(crisps)\n");printf(*2,爆米花(popcorn)\n");printf(*3,巧克力(chocolate)\n*);printf(*4J可樂(cola)\n");printf("5,退出(exit)\n");printf("****************************************\n*);while(i<=5)(printf("請輸入你要査詢的食品編號(hào)或者退出(査詢5次后將自動(dòng)退出)'n");scanf(*%d*,&c);switch(c)|printf("薯片(crisps)的單價(jià)為3.0元、n");break;printf("爆米花(popcorn)的單價(jià)為2.5元、n");break;printf("巧克カ(chocolate)的單價(jià)為4.0元、n");break;printf("可樂(cola)的單價(jià)為3.5元、n");break;printf("退出(exit)\n");break;default:printf("請輸入1?5之間的數(shù)字!'n");}i++；if(i=5)break;if(c=5) break;)continue;)elseprintf("\n\n歡迎使用自動(dòng)售賣系統(tǒng)ゝn");printf(ヽ***************************************、n");printf("1,薯片(crisps)\n");printf("2,爆米花(popcorn)\n");printf("3,巧克カ(chocolate)\n");printf("4?可樂(cola)\n");printf("5,退出(exit)\n");printf("****************************************\n");while(i<=5)(printf("請輸入你要査詢的食品編號(hào)或者退出、n");scanf("%d*,&c);switch(c)({printf("薯片(crisps)的單價(jià)為3.0元'n");printf("請輸入你想購買的數(shù)量'n");scanf("%d",&d);f=d*3;printf("請付詛id元'n”,f);printf(”請輸入您的付款金額:つ；scanf("%d",&e);if(e<f)(printf(“付款金額不足,請補(bǔ)交%d元'n",f-e);printf("請輸入您補(bǔ)交的金額つ;scanf("%d",&g);printf("找零%d元'n,g+e-f);}elseprintf("找零％d'元n",e-f);break;printf("爆米花(popcorn)的單價(jià)為2.5元'nつ;printf("請輸入你想購買的數(shù)量'nつ；scanfC*%d*,&d);f=d*3;printf("請付款%d元、n”,f);printf("請輸入您的付款金額:つ;scanf(*%d*,&e);if(e<f){printf("付款金額不足,請補(bǔ)交%d元、n”,f-e);printf("請輸入您補(bǔ)交的金額");scanf("%d",&g);printf("找零%d元、n*,g+e-f);}elseprintf(“找零%d元、n”,e-f);break;printf("巧克カ(chocolate)的單價(jià)為4.0元、n");printf("請輸入你想購買的數(shù)量、n");scanf("%d",&d);f=d*3;printf("請付款%d元、n”,f);printf("請輸入您的付款金額:つ;scanf("%d",&e);if(e<f){printf(“付款金額不足,請補(bǔ)交%d元、n",f-e);printf(〃請輸入您補(bǔ)交的金額づ;scanf("%d",&g);printf("找零%d元い",g+e-f);}elseprintf("找零%d元い",e-f);break;printf("可樂(cola)的單價(jià)為3.5元、nつ;printf("請輸入你想購買的數(shù)量、nつ;scanf("%d",&d);f=d*3;printf("請付動(dòng)d元、n",f);printf("請輸入您的付款金額:つ;scanf("%d",&e);if(e<f)(printf("付款金額不足,請補(bǔ)交%d元い",f-e);printf(〃請輸入您補(bǔ)交的金額づ;scanf("%d",&g);printf("找零%d元、n",g+e-f);elseprintf（"找零%d元、n”,e-f）;break;printf（"退出（exit）\n"）;break;default:printf（"請輸入1?5之間的數(shù)字!\n"）;}}break;}printf（=****************************************\n 謝謝惠顧!\n"）;return0;）}「c\C:\Docu>entsandSettings\AdMinistrator\^W\Debug\xq9.1?價(jià)格查詢系統(tǒng) 2.自動(dòng)售賣系統(tǒng)XXXXMXXXXX?メ"立エ丈"丈"メエM****X**>6XXXXXW請選擇您要逬行的操作:]歡迎使用價(jià)格查詢系統(tǒng)1,畫片(crisps)2t爆冬花(popcorn)3,巧克力《chocolate).可樂《cola)5,退出請輸入您要查詢的食品編號(hào)或者退出:(查詢5次后將自動(dòng)退出〉拿片的價(jià)格為3.。元番輸入也要查詢的食品編號(hào)或者退出:(查詢5次后將自動(dòng)退出〉1.價(jià)格查詢系統(tǒng) 2,自動(dòng)售賣系統(tǒng)請選擇您要逬行的操作:2川C川C一七ワC4*<帀ナく?宀片義樂出jf養(yǎng)巧可退i-M,,,,,4X12345*射*****請輸入您要購買的食品編號(hào)或者退出:常片的伊轡為3.0元蓄輸入徐栢購買的數(shù)量:*C:\Docu>entsandSettings\Adainist片輸3.買

知財(cái)18您足您.0費(fèi)不入12片輸3.買

知財(cái)18您足您.0費(fèi)不入12,ネ金春

付委206、本周小結(jié)通過本周的學(xué)習(xí),我對(duì)于switch語句已經(jīng)有了初步的認(rèn)識(shí)。我覺得switch語句是一個(gè)很好的東西,它就像ー個(gè)多項(xiàng)開關(guān),可以打開任何你想打開的語句段。在編寫最后ー題的代碼時(shí),我也用了很長的時(shí)間來寫。雖然說是編寫的很累,但是呢，我感受到了編寫完程序時(shí)的成就感和編寫時(shí)候的樂趣。我想,在以后的學(xué)習(xí)道路上,我會(huì)更加地努力去學(xué)習(xí)編寫代碼的。本次課程進(jìn)行課后習(xí)題測試,先作完的先遞交（我們不但考察質(zhì)量,還要兼顧速度（時(shí)間90分鐘））1、教材P129—-6.2;#include<stdio.h>intmain(){inti=l,m=l,n=l,1=1;chara;printf("請輸入ー組字符,并以enter結(jié)束:\n");while((a=getchar())!=*\n)(if(a>='?！?&a<='9')i++；elseif(a>='a'&&aく'z'||a>='A'&&a<='Z')m++;elseif(a=='')n++;else1++；)printf("字母的個(gè)數(shù)為％d\n",mT);printf("空格的個(gè)數(shù)為%d\rT,n-l);printf("數(shù)字的個(gè)數(shù)為％d\n",i-1);printf("其他字符的個(gè)數(shù)為%d\n",1-1);return0;し、 し:、ピnisaiiuごビ11丄Kigs胃6リ入ー歸&123人ス$ hjhを母的個(gè)數(shù)為3生將的個(gè)數(shù)為3數(shù)學(xué)的個(gè)數(shù)為3其他宗符的卜敦為3■Pressanykeytocontinue2、教材pl29--6.3;#includeくstdio.h>intmain()(inta,n,s=0,i=l,m=0;printf("請輸入a與n的值、n");scanf("%d%d”,&a,&n);while(i<=n)(m=m+a;s=s-hn;a=a*10;i++；)printf("a+aa+aaa+??,Gd\n,s);return0;}WSXa與n琉25a+aa+aaa+...=24690Pressanykeytocontinue3、教材pl29 6.4;#include<stdio.h>intmainOinti;floata=l,s=0;for(i=l;i<=20;i++)a=a*i;s=s+a;)printf("1!+2!+.??=%lf\n*,s);return0;)..=2561327455189073900.000000Pressanykeytocontinue.4、教材pl29 6.6;#includeくstdio.h>intmain()inti,j,k;printf("水仙數(shù)為:\n");for(i=l;i<=9;i++)for(j=0;j<=9;j++)for(k=0;k<=9;k++){if(i*i*i+j*j*j+k*k*k==100*i+10*j+k)printf("%d%d%d\n",i,j,k);)return0;}5、教材pl29一一6.10;#include<stdio.h>intmain()inti,j=l;for(i=l;i<10;i++)j=2*(j+l);printf("第一天共摘%d個(gè)桃子.\n”,j);return0;僅一天共摘1534下桃子.Pressanykeytocontinue1、使用數(shù)組完成123456789序列的逆序輸出#include<stdio.h>intmain()(inti,a[10];printf(〃原數(shù)列為:つ;for(i=0;i<10;i++)printf("a[%d]=%-2d",i+1,i);printf("\n處理后為:");for(i=9;i>=0;i―)printf("a[%d]=%-2d",9-i,i);printf('\n");return0;}原數(shù)列為:a[2]=la[3]=2a[4]=3a[5]=4at6]=5a[?]=6a[8]=7a[9]=8a[10]=9處理后為:a[0]=9atl]=8a[2J=7a13]=6a[4]=5a[5]=4a[6]=3a[?J=2a[8]=la[9]=0Pressanykeytocontinue.2、使用數(shù)組求Fibonacci數(shù)列的前20項(xiàng)。要求4項(xiàng)一行輸出。#includeくstdio.h>intmain()(inti,a[20]={l,1};printf("Fibonacci數(shù)列的前20項(xiàng)為:");for(i=2;i<20;i++)a[i]=a[i-2]+a[i-l];for(i=0;i<20;i++)(if(i%5==0)printf[\n");printf("\n");return0;Fihonacci數(shù)列的前20項(xiàng)為:112358132134558914423337761098715972584418167651Pi'essanykeytocontinue3、用數(shù)組編寫青年歌手大獎(jiǎng)賽的得分程序(要求10位評(píng)委,評(píng)委的給分必須位于(1.0-10.0)之間,然后去掉ー個(gè)最高分,再去掉ー個(gè)最低分,求歌手的最后得分(保留小數(shù)點(diǎn)后兩位數(shù))#include<stdio.h>intmain()(floata[10],min,max,sum=0,ave=0;inti;printfC請給分,分?jǐn)?shù)為1.0-10.0之間的數(shù)、rT);for(i=0;i<10;i++)(scanf("%f”,&a[i]);sum=sum+a[i];}min=a[0];for(i=0;i<10;i++)(if(min>a[i])min=a[i];}max二a[〇];for(i=0;i<10;i++)if(maxくa[i])max=a[i];}ave=(sum-max-min)/8;printf("最髙得分=%.2f,最低得分=%.2f,最后平均分=%.2f\n",max,min,ave);return0;4.請完成下列圖形的功能:#include<stdio.h>#deHneSIZE100intmain()(inta[SIZE],i,n,m,k=0;printf,請輸入數(shù)組元素的個(gè)數(shù):\n")；scanf(H%dM,&n);prinHV請輸入數(shù)組元素、n”)；for(i=0;l<n;i++)scanf(M%d,\&a[i]);printf("輸入査找的數(shù):")；scanf(H%dM,&m);for(i=0;i<n;i++){if(a[i]==m)(printf("此數(shù)位于數(shù)組中的序號(hào):％d\n”,i);k++;))if(k==0)printf("此數(shù)組中沒有％(1這個(gè)數(shù)、n”,m);elseprintf("此數(shù)組中有％d個(gè)%d\n]k,m);return0;1、使用二維數(shù)組完成3*3階矩陣的操作(主(副)對(duì)角線之和、形成上(下)三角矩陣與單位矩陣)ttinclude<stdio.h>voidmain()(inta[3][3]={1,3,5,7,9,11,13,15,17);intsum=0,i,j;printf("輸入的數(shù)組為:\n*);for(i=0;i<3;i++)(for(j=0;j<3;j++)printf(*%r4d*,a[i][j]);printf("\n");)printf("\n");for(i=0;i<3;i++)for(j=0;j<3;j++)if(j=i)sum=sum+a[i][j];printf(*+sum=%d\n*,stun);sum=0;for(i=0;i<3;i++)for(j=0;j<3;j++)if(j+i=2)sum=sum+a[i][j];printf[-sum=%d\n*,sum);printf('\n");for(i=0;i<3;i++)|for(j=0;j<3;j++)if(i>j)printf(*0 *,a[i][j]);elseprintfC%-4d”,a[i][j]);printf("\n");}printf("\n");for(i=0;i<3;i++)(for(j=0;j<3;j++)if(i<j)printf(*0 *,a[i][j]);elseprintfC%-4d*,a[i][j]);printfC\nO;}printf('\n");for(i=0;i<3;i++)|for(j=0;j<3;j++)if(i!=j)printf(*0 *,a[i][j]);elseprintf(*1 *,a[i][j]);printf(*\n*);2、教材P138例題7.4#include<stdio.h>intmainO(inta[2][3]={{l,2,3},{4,5,6}):intb[3][2],i,j;printf("數(shù)組a為:\n");for(i=0;i<=l;i++)(for(j=0;j<=2;j++)(printf("%5d",a[i][j]);b[j][i]=a[i][j]：}printf(*\n*);)printf("數(shù)組b為:\n")；for(i=0;i<=2;i++)(for(j=0;j<=l;j++)printf(,%5d*,b[i][j]);printf(*\n*);)return0;じ?、?し??U?し二ERUE?し數(shù)組ぶ:TOC\o"1-5"\h\z12 35 6數(shù)組b為:456Pressanykeytocontinue.3、教材P138例題7.5#includeくstdio.h>intmainO(inti,j,row=0,colum=0,max;inta[3][4]={{1,2,3,4,1,{9,8,7,6},{-10,10,-5,2});max=a[0][0];for(i=0;i<=2;i++)for(j=0;j<=3;j++)if(a[i][j]>max){max=a[i][j];row=i;colum=j;}printf(*max=%d,row=%d,cloum=^d\n*,max,row,colum);return0;iax=10,row=2,cloun=liessanykeytocontinue4、教材P151例題7.8#include<stdio.h>voidmainO(charstring[81];inti,num=0,word=0;charc;gets(string);for(i=0;(c=string[i])!='、〇';i++)if(c='9)word=0;elseif(word==0)(word=l;num++;}printf("字母的個(gè)數(shù)為％d\n",nu^l);}レonaremyfriends孕母的個(gè)數(shù)為4pressanykeytocontinue1、完成如ド圖形操作的編程#include<stdio.h>#defineSIZE100voidmain()inta[SIZE],i,11,111,k=0;printf(ハ請輸入數(shù)組元素的個(gè)數(shù):\nハ)；scanf(M%dM,&n);printf(ハ請輸入數(shù)組元素、曖)；for(i=0;i<n;i++)scanf(M%d'\&a[i]);printf("輸入査找的數(shù):scanf(M%d'\&m);for(i=0;i<n;i++)(if(a[i]==m)(printf("此數(shù)位于數(shù)組中的序號(hào):％d\n”,i);k++;})if(k==0)printf("此數(shù)組中沒有％d這個(gè)數(shù)、n”,m);elseprintf("此數(shù)組中有％d個(gè)%d\n",k,m);)2、已有宏定義SIZE為100,定義ー個(gè)整型數(shù)組a,數(shù)組元素個(gè)數(shù)n及數(shù)組元素從鍵盤輸入,對(duì)輸入的元素從大到小進(jìn)行排序,最后輸出排序后的數(shù)組。(選擇排序)、#include<stdio.h>#deHneSIZE100voidmain(){inta[SIZE],iJ,n,k,temp;printf("請輸入數(shù)組元素的個(gè)數(shù):つ;scanf(M%dM,&n);printf("請輸入數(shù)組元素:\nu);for(i=0;i<n;i++)scanf(,'%dM,&a(i]);for(i=0;i<=n-l;i++)(k=i;forQ=i+l;j<=n-l;j++)if(alj]>a[k])k=j；if(k!=i)1temp=a[i];a[i]=a[k];a[k]=temp;)}printf("從大到小排序后的數(shù)組為:\n");for(i=0;i<n;i++)請輸入數(shù)組元重的個(gè)數(shù):5請輸入數(shù)組元素:34234567從大到小排序后的數(shù)組為:67 45 34 23 23Pressanykeytocontinue3、完成如ド圖形操作的編程#include<stdio.h>#deHneSIZE100voidmain(){inta[SIZE]9num,x,n,i;printf「請輸入數(shù)組元素的個(gè)數(shù):\iT);scanf(M%dH,&n);printf(”請輸入各數(shù)組元素:\n”)；for(i=0;i<n;i++)scanグ％d”，&a[i]);printf("請輸入插入的數(shù)據(jù)num:\nM);scanf(M%dM,&num);printf,請輸入插入的位置x:\nn);scanf(H%dM,&x);

for(i=n-l;i>=x;i)a[i+l]=a[i];a[x]=num;n++;printf("數(shù)組元素有n=%d\nn,n);printf「插入后新數(shù)組為:\n")；for(i=0;i<n;i++)printf(M%5dH,a[l]);printf(M\nM);}バ?C:\DOCUIEITSATOSETTI胸入數(shù)組元素麻磯:ト輸入各數(shù)組元素:日卜請輸入插入的數(shù)據(jù)num：傲阻無幸n=5請輸入插入的位置x：傲阻無幸n=5插入后新數(shù)組為:76 34 56 45 23Pressanリkeytocontinue4、已定義數(shù)組:afSIZEl:數(shù)組的個(gè)數(shù)n,各元素的數(shù)據(jù)從鍵盤輸入(無序),現(xiàn)要?jiǎng)h除某位置上的數(shù)值,其中刪除的位置從鍵盤輸入。#include<stdio.h>#deflneSIZE100voidniain()(inta[SIZE],x,n,i;printf,請輸入數(shù)組元素的個(gè)數(shù)、\nH);scanf(H%d'\&n);printf「請輸入數(shù)組的%d元素,元素間用空格\nハ,n);for(i=0;i<n;i++)scanグ％d”，&a|i]);do{printf("請輸入要?jiǎng)h除數(shù)據(jù)的位置:\n");scanf(M%dM,&x);if(x>nIIx<0)printf("輸入刪除的位置有錯(cuò)\11つ;elsebreak;}whiled);for(i=x-l;i<n-l;i++)a[i]=a[i+l];printf("刪除第%d元素后的新數(shù)組為\nk);for(i=0;i<n-l;i++)printグ％?5d”,a[i]);printf(M\nH);バ?C:\DOCU1ENTSANDSETTINGS\AD1IHIST請輸入數(shù)組元素界最雋輸入數(shù)組的5個(gè)元素,元素間用空格情輸入要?jiǎng)h除數(shù)據(jù)的位置:的除第。個(gè)元素后的新數(shù)組為Pressanykeytocontinue5、完成如下圖示的操作#include<stdio.h>intmain()(intiJ,k,m,n,a[1000];printf("請輸入數(shù)組的長度八n”)；scanf(H%dH,&m);printf("原數(shù)組為:\n”)；for(i=0;i<m;l++)scanf(H%d'\&a[i]);while(l){printf,請選擇:1:刪除偶數(shù)后的數(shù)組2:刪除ー個(gè)數(shù)后的數(shù)組3:退出\n");scanf(H%d'\&n);if(n==3)break;if(n==l)printf（"刪除偶數(shù)后的數(shù)組為:\n"）;for(i=0;i<m;i++)if(a[i]%2!=0)printf(M%-4dH,a[i]);if(n==2){printグ請輸入要?jiǎng)h除的數(shù)八n")；scanf(H%dM,&k);printf("刪除％d后的數(shù)組為:\n*\k);for(i=0;i<m;i++)if(a[i]!=k)printグ％?4d”,a[i]);)printf(H\nM);}return0;や輸入數(shù)組的長度:る數(shù)組知請選擇:1:刪除偶數(shù)后的數(shù)組2:刪除ー個(gè)數(shù)后的數(shù)組3:退出刪除偶數(shù)后的數(shù)組為:請選擇:1:刪除偶數(shù)后的數(shù)組2:刪除ー個(gè)數(shù)后的數(shù)組3:退出請輸入要?jiǎng)h除的數(shù):刪除5后的數(shù)組為:請選擇:1:刪除偶數(shù)后的數(shù)組2:刪除ー個(gè)數(shù)后的數(shù)組3:退出Pressanykeytocontinue.1012141618201320013459刪除偶數(shù)后的數(shù)組為:1359Pressanykeytocontinue原數(shù)組為:,除87后的新數(shù)組為2 3 20 58 72 76 78 90 900Pressanykeytocontinue.6、對(duì)前ー階段C語言學(xué)習(xí)的總結(jié)(200-300字)C程序的題目真得很有意思,可以完成自己想要的程序。有時(shí)候題目很難,這個(gè)時(shí)候,可以去問問同學(xué),借借代碼。這樣和同學(xué)的交流多了,也使得和同學(xué)的關(guān)系更加融洽。函數(shù)作業(yè)1、使用用戶函數(shù),求任意輸入四個(gè)整數(shù)的最大值#include<stdio.h>intmain()(inta9b9c9d9e;intmax(intx,inty);scanf(M%d,%d,%d,%dH,&a,&b,&c,&d);e=max(max(a,b),max(c,d));printf(Mmaxis%d\nM,e);return0;}intmax(intxjnty)]intz;if(x>y)z=x;elsez=y;return(z);}レ5,6〃)axis7anykeytocontinue.使用用戶函數(shù),求任意輸入ー個(gè)整數(shù)的階乘#include<stdio.h>intmain()(inta,b;intl*act(intx);scanf「％dj,&a);b=fact(a);printf("factis%d\n0,b);return0;}intfact(intx)intz=l,i；for(i=l;i<=x;i++)z=z*i;return(z);}Bactis120ressanykeytocontinue(1)定義函數(shù)fact(n)計(jì)算n的階乘:n!=1*2* * n?函數(shù)類型是double。(2)定義函數(shù)cal(m,n)計(jì)算累加和:s=gk=m+(m+l)+ +〃,函數(shù)類型是double。k=m(3)定義函數(shù)main。,輸入正整數(shù)n,計(jì)算并輸出下列算式的值。該算式中,每ー項(xiàng)的分子是累加和,要求調(diào)用函數(shù)cal(m,n)計(jì)算ナム:每ー項(xiàng)的分母是階乘,要求調(diào)用函數(shù)k-mfact(n)計(jì)算n!〇#include<stdio.h>doublemain()(doublefact(doublex);doublecaKdoublem,doublen);doublesum=0;intn,i;scanf(M%du,&n);for(i=l;i<=n;i++)sum=sum+(cal(lj))/(fact(i));printf(Hsum=%lf\nM,sum);return0;)doublefact(doublex){doublez=l,i;for(i=l;i<=x;i++)z=z*i;return(z);doublecaKdoublem,doublen)(doublez=0,i;for(i=m;i<=n;i++)z=z+i;return(z);}定義函數(shù)fact(n)計(jì)算n的階乘:n!=1*2*……*n,函數(shù)返回值類型是double。定義函數(shù)cal(e)計(jì)算下列算式的值,直到最后ー項(xiàng)的絕對(duì)值小于。,函數(shù)返回值類型是double。.111S=1+——+-4--+.….2!3!4!(3)定義函數(shù)main。,輸入正整數(shù)n,當(dāng)精度e分別取值為\,、10\70\……ヽノ°"時(shí),分別計(jì)算并輸出下列算式的值,直到最后一項(xiàng)的絕對(duì)值小于精度e,以比較不同精度下算出的結(jié)果。要求調(diào)用函數(shù)cal(e)計(jì)算下列算式的值。111S=1+—+—I 1-2!3!4!#include<stdio.h>doublemain()(doublefact(doublex);doublecaKdoublee);doubles=0;intn;scanf(H%dH,&n);s=cal(e)；printf(Hs=%lf\nH,s);return0;}doublefact(doublex)doublez=l,i;for(i=l;i<=x;i++)z=z*i;return(z);}doublecaKdoublee)(doublez=0,i;for(i=l;fact(i)<=(l/e);i++)z=z+l/fact(i);return(z);}#include<stdio.h>#inciude<stdlib.h>#include<math.h>voidrunyear(){inta;printグ請輸入201I至3000間任意年份ハ慶)；scanf(H%dH,&a);if(a%1000==0)if(a%400==0)printf「％d是閏年Xn[a);elseprint*"%d不是閏年\n]a);elseif(a%4==0&&a%100!=0)printf(”％d是閏年Xn\a);elseprintf(H%d不是閏年Xn”,a);}voidfc(){floata,b,c,d,xl,x2,x;printグ請輸入a,b,c的值(并以空格隔開)\n”);scanf(M%f%f%fM,&a,&b,&c);d=b*b-4*a*c;if(d<0)printf("該方程無解、n")；elseif(d>0)xl=(-b+sqrt(d))/(2*a);x2=(-b-sqrt(d))/(2*a);printf("%.2fxA2+%.2fx+%.2f=0^^^\nxl=%.2f\nx2=%.2An",a,b,c,xl,x2);)else{x=(-b+sqrt(d))/(2*a);printff"%2fx"2+%.2fx+%.2f=0的唯一解為\nx=%.2An",a,b,c,x);voidfact()(inti=lja,s=l;printf("請輸入一個(gè)小于6的正整數(shù)、n");scanf(n%dM,&a);while(i<=a)(s=s*i;i++；)printf(ir%d!=%d\n'\a,s);)voidcfbO{intij,k,m;for(k=l;k<10;k++){printf(M%8d'\k);)printf(n\nn);for(k=l;k<10;k++)printf(M M);printf(M\nn);for(i=l;i<10;i++)(for(j=l;j<=i;j++)printf(M%2d*%d=％-3dn,i,j,i*j);printf('ヽ］'1);voidstar()

intij,k;for(i=4;i<5,i>=(-l);i-)for(j=0；j<=i+l；j++)printグ")；for(k=0;k<4-i;k++)printf(H*H);printf”iT);)for(i=4;i<5,i>=l;i?)(for(j=0;j<6-i;j++)printf(Mu);for(k=O;k<=i-l;k++)printf(M*M);printf(M\nM);voidShowMenuOwhile(l)intinput;charch;printf("\n");printf(HKprintf(“2、printf(“3、printf(HKprintf(“2、printf(“3、printf(H4.printf(“5、printf(M6>求ax2+bx+c=0方程的根'n"）;任意輸入一整數(shù)（小于6）,求其階乘'ザ）；九九乘法表'n"）；輸出寶石星'n”）；退出'n"）；printf(''**********************************************\n'').scanf(M%dH,&input);switch(input)runyear();break;fc()；break;fact();break;cfb();break;star();break;exit(O);Iprintfむ輸入Y或者、繼續(xù),輸入其它字符退出");getcharO;scanf(M%cM,&ch);if(ch==,Y,llch==,y,)continue;elsebreak;)}intmain()ShowMenuO;return0;}?*MMM??XXMM M關(guān)?興?共*-X輸入丫或者y繼續(xù),輸入其它字符退出y判斷某一年是否閏年［2011-3000之間）求ax2+bx*c=0方程的根任意題入一整數(shù)（小于6）,求其階乘九九無法表輸出寶石星求二維數(shù)組最大值#include<stdio.h>intmax_value(intarray[][4])(intij,max;max=array[O][O];for(i=0;i<3;i++)for(j=0;j<4;j++)if(array[i][j]>max)max=array[i][j];return(max);)intmain(){inta[3][4]={{l,3,5,7},{2,4,6,8},{15,100,34,12}};printf(Mmax_valueis%d\nH,max_value(a));retuni0;iax_ualueis100ressanykeytocontinu8.11,求數(shù)組的平均值#include<stdio.h>voidmain(){inti;floataverage(floatarray[10]);floatscore|lOliver;for(i=0;i<10;i++)scanf(H%f,,,&score[i]);printf(H\nH);aver=average(score);printf(M%.2f\nM,aver);}floataverage(floatarray!1()])(inti;floataver,sum=0;for(i=0;i<10;i++)sum+=array[i];aver=sum/10;returnaver;}J23456789106.50[Pressanykeytocontinue8.13,排序#include<stdio.h>voidmain()(voidsort(intarrayロ,intn);inta[10],i;printf("請輸入ー個(gè)數(shù)組:\nM);for(i=0;i<10;i++)scanf(''%(T',&a[i]);sort(a,10);printf(”被排序出的數(shù)組是:'n”);for(i=0;i<10;i++)printf(M%5dM,a[i]);printf(H\nn);)voidsort(intarray[],intn)(intij,k,t；for(i=0;i<n-l;i++){k=i;for(j=i+l;j<ny++)if(array[j]<array[k])k=j；t=array[k];array[k]=array[i];array[i]=t;)}請輸入一個(gè)數(shù)組:5645782354被排序出的數(shù)組是:2 3 4 4 5Pressanykeytocontinue本次作業(yè)是代碼分析,排版好程序代碼的編寫風(fēng)格,并對(duì)函數(shù)做說明(注解)然后調(diào)試出結(jié)果,學(xué)出調(diào)試后的代碼實(shí)現(xiàn)的含義#include<stdio.h>#include<stdlib.h>float fac(intn)〃定義ー個(gè)遞歸函數(shù)求階乘(floatf;〃定義變量fif(n<=0)〃排除錯(cuò)誤的情況{printf("n<0,DataError!\n\n");exit(O);}elseif(n==l)f=l；elsef=fac(n-l)*n;return(f);}intmain。〃主函數(shù)(do(intn;floaty;printfC'InputaIntegerNumber:**);scanf(**%d**,&n);y=fac(n);printf(**%d!=%0.0f\n\n**,n,y);}while(l);return0;}本代碼是通過定義遞歸函數(shù)實(shí)現(xiàn)求整數(shù)階乘的功能。#include<stdio.h>intsortfintb[],intn,intx)〃定義ー個(gè)遞歸函數(shù),比較數(shù)組中的相鄰兩個(gè)數(shù),并用小數(shù)取代大數(shù).(intk;for(k=n-l;k>=0;k-)if(b[k]<x)b[k+l]=b[k];elsebreak;b[k+l]=x;return0;}intmain()(intij;staticinta[5]={8,5,4,l,6};intb[8];b[O]=a[OJ;for(i=l;i<5;i++)(sort(b,i,a[i]);for(j=0;j<=i;j++)printf(**%d**,b[i]);printf(**\n**);return0;}〃利用已經(jīng)定義的遞歸函數(shù),求數(shù)組中最小的3個(gè)數(shù),并進(jìn)行排列#include<stdio.h>intfun(intx)〃定義ー個(gè)遞歸函數(shù)求一個(gè)分段代數(shù)函數(shù)(intp;if(x==Ollx==l)return3;elsep=x-fun(x-2);returnp;}intmain()(printf(n\n%d\nM,fun(9));return0;}〃本函數(shù)主要實(shí)現(xiàn)了當(dāng)x=9時(shí)的分段函數(shù)求解#include<stdio.h>intmain()〃主函數(shù)(intf(int);f(1234);return0;}intf(intn)〃定義ー個(gè)函數(shù),將n的每個(gè)位數(shù)上的數(shù)字分開if(n>=10)f(n/10);printf(H%d\nH,n%10);return0;}〃定義ー個(gè)函數(shù),將n的每個(gè)位數(shù)上的數(shù)字分開#include<stdio.h>intsum(intn)〃定義ー個(gè)遞歸函數(shù),求s=l+l+2+l+2+3+l+ n的值(intp=O,s=OJ;for(j=l;j<=n;j++)s+=(p+=j);returns;}intmain()〃主函數(shù){printf(Msum(5)=%d\n*\sum(5));return0;}〃求s=l+l+2+l+2+3+l+ n的值#include<stdio.h>intsum(intn)〃定義ー個(gè)遞歸函數(shù),求ー個(gè)整數(shù)的各位數(shù)數(shù)字之和{intc=0;do{c+=n%10;n/=10;}while(n);returnc;)intmain(){printf(Hsum(123)=%d\nH,sum(123));return0;}〃定義ー個(gè)遞歸函數(shù),求一個(gè)整數(shù)的各位數(shù)數(shù)字之和#include<stdio.h>intfun(intn)〃求s=n+(n?2)+…+0之和if(n>0)return(n+fun(n-2));elsereturn0;}intmain()(intn=10;intfun(intn);printf(H%d\nH,fun(n));return0;}〃求s=n+(n-2)+...+0之和本次作業(yè)是教材上的習(xí)題,主要是測試使用函數(shù)處理問題的題目P202…習(xí)題8.1#include<stdio.h>voidniain()(inthcf(int,int);intlcd(int,int,iiit);intu,v,h,l;scanf("%d,%d,,,&u,&v);h=hcf(u,v);printf("HCF=%d\n”,h);〃最大公約數(shù)l=lcd(u,v,h);printf("LCD=%d\n"J);〃最小公倍數(shù))inthcf(intujntv){intt,r;if(v>u)|t=u;u=v;v=t;}while((r==u%v)!=0)(u=v;v=r;)return(v);}intlcd(intu,intv,inth)(return(u*v/h);}P202…習(xí)題8.2#include<stdio.h>#include<math.h>floatxl,x2,disc,p,q;voidmain()voidgreater_than_zero(float,float);voidequal_to_zero(float,float);voidsmaller_than_zero(float,float);floata,b,c;printf("\ninputa,b,c;H);scanf(M%f,%f,%fH,&a,&b,&c);printf(''equation:%5.2f*x*x+%5.2f*x*x+%5.2f=0\n",a,b,c);disc=b*b-4*a*c;printf("root:\n");if(disc>0)(greater_than_zero(a,b);printf("xl=%f\t\tx2=%f\n",xl,x2);)elseif(disc==0)(equal_to_zero(a,b);printf("xl=%f\t\tx2=%f\n",xl,x2);elsesmaller_than_zero(a,b);printf("xl=%f+%fi\tx2=%f-%fi\n",p,q,p,q);voidgreater_than_zero(floata,floatb)xl=(-b+sqrt(disc))/(2*a);x2=(-b-sqrt(disc))/(2*a);voidequal_to_zero(floata,floatb)xl=x2=(-b)/(2*a);voidsmaller_than_zero(floata,floatb)p=-b/(2*a);q=sqrt(-disc)/(2*a);P202…習(xí)題8.3#inchide<stdio.h>voidmain()(intprime(int);intn;printf(M\ninputaninteger:H);scanf(M%dH,&n);if(prime(n))printf(',\n%disaprime.\nM,n);elseprintf(M\n%disnotaprime.\nH,n);1intprime(intn)(intflag=l,i;for(i=2;i<n/2&&flag==l;i+4-)if(n%i==0)flag=0;return(flag);)P202一習(xí)題8.5#include<stdio.h>#include<string.h>voidmain()(voidinverse(charstr[]);charstr[1OO];printf(''inputstring:")；scanf(ハ％s”Mr);inverse(str);printf(Hinversestring:%s\nH,str);)voidinverse(charstr[]){chart;intij;for(i=0J=strlen(str);i<(strlen(str)/2);i++j-)t=str[i];str[i]=str[j-i];str[j-l]=t;P202一習(xí)題8.10#include<stdio.h>#include<string.h>voidmain(){intalphabetic(char);intlongest(char[]);inti;charline[100];printfC'inputoneline:\nH);gets(line);printf(HThelongestwordis:H);for(i=longest(line);alphabetic(line|i]);i++)printf(M%cM,line[i]);printf(u\nM);}intalphabetic(charc)(if((c>='a,&&c<=,z*)ll(c>=,A,&&c<=,z,))return1;elsereturn0;}intlongest(charstring[]){intlen=0,i,length=0,flag=l,place=0,point;for(i=0;i<=strlen(string);i++)if(alphabetic(string[i]))if(nag)(point=i;flag=0;}elselen++;else(flag=l;if(len>=length){length=len;place=point;len=O;))return(place);}P202一習(xí)題8.11#include<stdio.h>#include<string.h>#defineN10charstr[N];voidmain()(voidsort(char[J);intijlag;for(flag=l;flag==l;)(printf(Minputstring:\nM);scanf(H%sH,&str);if(strlen(str)>N)printf(Mstringtoolongjnputagain!");elseflag=0;)sort(str);printf("stringsorted:\n");for(i=0;i<N;i++)printf("%c"^tr[i]);printf("\n");}voidsort(charstr[]){intij;chart;for(i=l;j<N;j++)for(i=0;(i<N-j)&&(str[i]!=,\0,);i++)if(str[i]>str[i+l])t=str[i];str[i]=str[i+l];str[i+l]=t;本次作業(yè)是教材上的例題與習(xí)題,主要是測試預(yù)處理問題的題目P205-一例題9.1#include<stdio.h>#deHnePI3.1415926voidmain(){floatprintf("inputradius:")；scanf(H%F\&r);1=2.〇辛Pドr;s=PI*r*r;v=4.0/3*Pl*r*r*r;printf(Ml=%10.4f\ns=%10.4Anv=%10.4f\n"ノ舟v);)P210…例題9.5#include<stdio.h>#definePRprintf#defineNLH\nM#defineDM%dM#deHneDIDNL#defineD2DDNL#defineD3DDDNL#defineD4DDDDNL#defineSH%sHvoidmain(){inta,b,c,d;charstring[]=HCHINAH;a=l;b=2;c=3;d=4;PR(Dl,a);PR(D2,a,b);PR(D3,a,b,c);PR(D4,a,b,c,d)；PR(S,string);PR(NL);}P216…例題9.7#include<stdio.h>#deHneLETTER〇voidmain(){charstr[20]=uCLanguage”4;inti;i=0;while((c=str[i])!=*\O'){1++；#ifLETTERif(c>='a'&&cv='z')c=c-32;#elseif(c>='A'&&c<='Z')c=c+32;#endifprintf("%c",c);)printf("\n");)P217…習(xí)題9.2#include<stdio.h>#defmeSURPLUS(a,b)?a)%(b))voidmain(){inta,b;printf("inputtwointegera,b:");scanf("%d,%d",&a,&b);printf("remainderis%d\n",SURPLUS(a,b));}_P218一習(xí)題9.3#include<stdio.h>#include<math.h>#defineS(a,b,c)(a+b+c)/2#defineAREA(a,b,c)sqrt(S(a,b,c)*(S(a,b,c)-a)*(S(a,b,c)-b)*(S(a,b,c)-c))voidmain()(floata,b,c;printf("inputa,b,c:");scanf("%f,%f,%f",&a,&b,&c);if(a+b>c&&a+c>b&&b+c>a)printf("area:%8.2f\n"AREA(a,b,c));elseprintf("isnotatriangle!");)P218…習(xí)題9.4#include<stdio.h>#defmeLEAP_YEAR(y)((y%4==0)&&(y%100!=0)ll(y%400==0))voidmain()intyear;printf(H\ninputyear:\nM);scanf(1,%dM,&year);lf(LEAP_YEAR(year))printf(**%disaleapyear\nM,year);elseprintf(**%disnotaleapyear\n'\year);}本次作業(yè)主要是測試大家對(duì)新概念的掌握1、寫出結(jié)構(gòu)體定義的格式(含有8個(gè)成員項(xiàng))structstudent(intnum;charname[20];charsex;intage;floatscore;charaddr[30];longcellphone;longtelephone;{studentl,stufent2;.寫出結(jié)構(gòu)體數(shù)組(數(shù)組含有4個(gè)元素)定義的格式(含有4個(gè)成員項(xiàng))structstudent{intnum;charname[20];charsex;intage;}stu[4J;.理解與概念給出的代碼(要求代碼具有統(tǒng)計(jì)廢票功能)#include<stdio.h>/*I/O函數(shù)*/#include〈string.h>/?字符串函數(shù)?/structpersoncharname[20];intcount;｝；intmain(void)intk,t;charl_name[20];structpersonlea加r[4]=ドA'OJ'B”,〇,”C",0JD、。};printグ請給A、B和C三個(gè)人投票:\n“)；for(k=l;k<=10;k++)(scanf(M%sM,I_name);for(t=0;t<4;t++)if(strcmp(Lname,leader[t].name)==O)leader[t].count++;)printf("\n投票結(jié)果是:'n")；for(t=0;t<3;t++)printf(M%10s:%5dM"9leader[t].name,leader[t].count);printf(H\n廢票有:％5d票"，Ieader[3].counり;printf(n\nM);return0;}4 P318.H.3#include<stdio.h>#defineN5structstudent(charnum[6];charname|8];intscore[4];}stu[N];voidmain()voidprint(structstudentstu[6]);intij;for(i=0;i<N;i++)printf(M\ninputscoreofstudent%d:\n”,i+l);printf(HNO.:H);scanf(H%sM^tu[i].num);printf(Hname:H);scanf(M%s*>tu[i].name);for(j=0;j<3;j++){printf(,'score%d:Mj+l);scanf(H%dn,&stu[i].score|j|);}printf(H\nM);}print(stu);)ー、閱讀程序,讀出結(jié)果與機(jī)器運(yùn)行結(jié)果比較#include<stdio.h>intmain()intx;int*p;p=&x;scanf(H%dM,&x);printf(',%d\nn,*p);return0;該程序的意思就是你輸入了a,程序就會(huì)輸出a?因?yàn)樵诔绦蜻\(yùn)行時(shí),系統(tǒng)把輸入的a的地址給了p?然后輸出*p,即是輸出a的值。#include<stdio.h>intswap(int*pl,int*p2){intp;p=*pl;*pl=*p2;*p2=p;return0;}intmain(){inta,b;int*pl,*p2;scanf("%d%d”,&a,&b);pl=&a;p2=&b;swap(pl,p2);printf("%d%d\n",a,b);return0;)該程序的意思是通過指針交換兩個(gè)數(shù)的值。#include<stdio.h>voidmain(){int*pl,x=10;float*p2,y=2.5;pl=&x;p2=&y;printf(n%d,%f\nM,++(*pl),(*p2)++)5 〃++(*pl)在輸出之前加該程序的意思通過指針,將X的地址賦給pl,將y的地址賦給p2,然后++(*pl)是在輸出之前加所以輸出的為11,而(*p2)++是輸出之后加的,所以輸出的是2.500000.#include<stdio.h>voidmain(){inta[]={l,2,3,4,5,6),*p=a;*(p+3)+=2;printf("%d,%d\n",*p,*(p+3));)該程序是數(shù)組和指針的組合，*p=a,就是說?p是數(shù)組aロ，輸出?p即是輸出數(shù)組第一項(xiàng),輸出?(p+3)即是輸出6由(4+2)而來5,#include<stdio.h>voidmain(){inta[]={l,2,3,4,5,6),*p;P=a；printf(***%d,\n",*p);printf("*%d,\n",*(++p));printf(',*%d,\n",*++p);printf(',*%d,\n",*(p-));p+=3;printf('**%d,%d\n,,,*p,*(a+3));}*p即是第一項(xiàng)1*(++p)為2*++p為3*(p--)為2后來p+=3;*p為5 *(a+3)為4二、使用指針計(jì)算數(shù)組元素之和#include<stdio.h>voidmain()(inta[10],*p,i,s=0;P=a；printf(Mpleaseinput10munbers:\nM);for(i=0;i<10;i++)scanf(M%d*\p++);printf(M\nM);P=a；for(i=0;i<10;i++,p++)s=s+*p;printfC'sumis%d.\nM,s);)pleaseinput10nunbers：L234456789sumis49.Pressanykeytocontinue.三、使用指針輸出數(shù)組中的全部元素。#