編程-計(jì)概a00-04分支循環(huán)例題

1 153=153= a=i/b=(i–a*100)/c=i%if((a*a*a+b*b*b+c*c*c)==}} {for(i=1;i<=9;i++)for(k=0;k<=9;cout<<i<<j<<k<<endl;}方法1和2都是900次} 求=aaa+aaa+┅+aa…a例如：a=4,n=5=4+44+444+4444+44444=444=4*100+4*101+4*x=0;for(j =0; <3;x=x+ *pow(10,j)55={for(j=0;j<i;}{}66{intx=for(j=0;j<i;{}} +x0;{x=x*10+a;}單重循環(huán)執(zhí)行次數(shù)為n次，雙重循環(huán)n2 {cin>>a>>n;{sum+=x;}} 求int{inti,j,n=1,sum=for(i=1;i<=10;{for(n=1,j=1;j<=i;j++)n=n*j;sum=sum+}

//計(jì)算icout<<return}99int{inti,j,n=1,sum=for(i=1;i<=10;{n=n*sum=sum+

}cout<<sum<<endl;return0;}

958597*speed int{ints,a,b,c,e,d;doublewhile(s<=99999 a=s/b=(s–a*10000)/c=(s–a*10000–b*1000)/d=(s–a*10000–b*1000–c*100)/e=s–a*10000–b*1000–c*100–d*f=s–if(a==e&&b==d&&f%7==0cout<<“Thespeedis”<<f/7<<}return intints,a,b,c,d,for(i=1;i<=400;{s=95859+i*a=s/b=(s–a*10000)/c=(s–a*10000–b*1000)/d=(s–a*10000–b*1000–c*100)/e=s%if(a==e&&cout<<“Thespeedis”<<i<<}return} {ints,i,for(i=6;i<=9;{for(j=0;j<=9; s=(90000+i*1000+j*100+i*10+9)-95859;if(s%7==0)cout<<“Thespeedis”<<s/7<<}（9（9-5）*10+ s=95959-95859;if(s%7==0)cout<<“Thespeedis”<<return}

/7<< {if(num==0) elseif((num%2)==0)sum+=num;}} numcountcount 753

int{intnum;intcnt=0;cout<<"Pleaseenteraninteger."<<endl;cin>>num;{cout<<num%10;num=num/10;}while(num!=0

為什么不是whilecout<<"total"<<cnt<<"digits"<<endl;return0;為什么不是while}**j=2*i-每多一行,少一個(gè)空格k8iwhile(i<=8){while(k<=8-while(j<=(2*i–}int{inti=1,j,{k

while(k(21i))//21 cout<<"";k++;}j=1;while(j<=(2*i- cout<<"*";j++;}cout<<endl;}return}int{for(inti=1;i<=8;{

intlen1=21-i;intlen2=2*i-1;cout<<setw(len1)<<setfill('')<<"cout<<setw(len2)<<setfill('*')<<"*"<<}return} 思路：先把數(shù)字當(dāng)作for(行數(shù){40m*3再輸出m個(gè)*,間隔為k；//k}

* for(m=1;m<=10;{

coutsetw(40m*3//每行縮進(jìn)3個(gè)空格{cout<<setw(6)<<“*”}

cout<<}

通項(xiàng)c,初值為c=c0*m–n+1)/(n 以第行為4=1*(5–2+1)/(2-6=4*(5–3+1)/(3– //cout<setw(40)<1<{c0=

{cc0*mn1n1)c0=c; }} for(i=for(i=2;i<=k;if((m%i)==if(i==k+1)//判斷循環(huán)是否正常結(jié)束cout<<m<<“isprimecout<<m<<“isnotaint{intm,i,for(m=100;m<=200;{k=for(i=2;i<=k;if((mi0)break//只要被除盡，退出if(i==k+1)//正常退出循環(huán)cout<<m<<“isprimenumber\n”;cout<<m<<“isnota}return} h=(b-a)/ Y0=Y0=sin(a)Yn=∫sinxdx=(y0+y1)*h/2+(y1+y2)*h/2+…(yn-1+=(y0+2*(y1+y2+y3+…yn-1+yn)-#include int{doublea,b,h,temp,sum;inti,cin>>a>>b;//輸入的單位是弧度cout“cin>>h=(b-a)/n;sum=sin(a);for(i=1;i<=n;{temp=fabs(sin(a+i*h))*h;sum=sum+2*temp;}

sin(b)=ynsum=(sum–sin(b))*h/2cout“return} 求

x4

4 f

f(x1)x4 f(x1 f x f(xn- f’(xn- int doublex,x0,f,x1.5;//猜一個(gè)初值 x0 f((2*x04*x03*x06;//代入x0算函數(shù)f(xf1=(6*x08*x03x0算求導(dǎo)以后函數(shù)f(x)x=x0-f/}while(fabs(x-x0)>1e-cout<<xendl;return0;}千萬(wàn)要留意，無論是float是double型的變量，都！=”與數(shù)字比較，if(x0.0含錯(cuò)誤的比constdoubledeta1e-6(10-6）（大小寫均可）constdoubledeta=10e-6(10-5）if((x>=-deta)&&(x<=deta))if(fabs(x)<=1e-6)其中deta是允許的誤差（即精度）也可以用絕對(duì)值的誤差做衡量標(biāo)準(zhǔn) firstYear=2；thirdYear=4；forthYead=？firstYear=thirdYear=forthyear;

int {firstYear=2;secondYear=3;thirdYear=4;for(i=4;i<=20;i++){forthYear=firstYear+thirdYear;firstYear=secondYear;secondYear=thirdYear;thirdYear=forthYear;}cout<<"20yearslater,thereare<<forthYear<<"cows"<<return}–例如： 迭代中的x=x0-f/f1;–例如： 迭代中的x0=x;[例]兩個(gè)乒乓球隊(duì)進(jìn)行比賽,各出三人.甲隊(duì)為A,B,C三人,乙隊(duì)為X,Y,ZA不和XC不和X,Z比,請(qǐng)排出三對(duì)賽手的對(duì)陣.試探（窮舉） for(a=‘X’,a<=‘Z’,cout<<“A→”<<int{charfor(a=‘x’;a<=‘z’;a++)if(a!=x)for(b=‘x’;b<=‘z’;{

if(b for(c=‘x’;c<={if((c!=a)&&(c!=b)(c‘x’&&(c‘z’//c不和x，zcout<<“A→”<<a<<“B→”<<b<<“C→”<<}}return}

AAB000101011111顯然這是或的關(guān)系，因此有CC1=(A||CC2=!(A&&AD10011110000101011111110110110111001011111101101101110010： ：CC3=(A&&E）||（A&&F）||（E&&F)

第一種情況：同時(shí)作案（B&&CC4=(B&&C)||(!B&&?1100101?1001?1100101?1001000?0110000?0011011CC5=(C&&!D)||(D&& 含 D作案，E也作可1011D作案，E不作可0011D不作案，E也不可能作0100D不作案，E卻作不可如果作案了，不管E如果作案了，不管E CC5=(C&&!D)||(D&&案CC6=D||!E CC=CC1&&CC2&&CC3&&CC4&&CC5&&for(A=0;A<=1;for(B=0;B<=1;B++)for(C=0;C<=1;for(D=0;D<=1;for(E=0;E<=1;E++)for(F=0;F<=1;{求出ifint for(A=0;A<=1;A++)for(B=0;B<=1;for(C=0;C<=1;C++)for(D=0;D<=1;D++)for(E=0;E<=1;for(F=0;F<=1;{CC1=A||CC2=(A&&E)||(A&&F)||(E&&F);CC3=!(A&&D);CC4=(B&&C)||(!B&&CC5=(C&&!D)||(D&&!C);CC6=D||!E; cout<<"A:"<<(A==0?"innocent":"guilty")<<endl;cout<<"B:"<<(B==0?"innocent":"guilty")<<end;cout<<"C:"<<(C==0?"innocent":"guilty")<<endl;cout<<"D:"<<(D==0?"innocent":"guilty")<<endl;cout<<"E:"<<(E==0?"innocent":"guilty")<<endl;cout<<"F:"<<(F==0?"innocent":"guilty")<<endl;return0;}A選手說：B==2,A==3；B選手說：B==2,E==4；C選手說：C==1,D==2；D選手說：C==5D==3；E選手說：E==4,A==1； 2考慮到每個(gè)人說的話是一對(duì)一錯(cuò)，即一真一假，比如Ata=((B==2)+(A==3))==1;符合A選手的話，則ta為1tb=((B==2)+(E==4))==1;符合B選手的話，則tb為1tc=((C==1)+(D==2))==1;符合C選手的話，則tc為1td=((C==5)+(D==3))==1符合D選手的話，則td為1te=((E==4)+(A==1))==1;符合E選手的話，則te為10tatbtctdtefor(A=1;A<=5;A++){if{if(C==A||

A,B,C,D,E{1,2,3,4,5}ABCD{if(D==A||D==B||D==C)continue;for（E=1;E<=5;E++){if(E==D||E==A||E==B||E==C)A,B,C,D,E{1,2,3,4,5}ABCDE=15-(A+B+C+D)A*B*C*D*E==for(A=1;A=5; =1;B=5;B= =1;C=5;C++)for(D=1;D=5;E=15-(A+B+C+D); A*B*C*D*E== ta=((B==2)+(A==3))==tb=((B==2)+(E==4))==tc=((C==1)+(D==2))==td=((C==5)+(D==3))==te=((E==4)+(A==1))==t=ta+tb+tc+td+ t== 輸 x+y+z=5*x+3*y+z/3=x的范圍0~20，y的范圍0~33，z100x limx（其中x為任意常數(shù)，a為常數(shù) <= 再驗(yàn)證y=n–x是不是素?cái)?shù)。 for(x=3,x<=n/2,x+=如果xynx比如：q/p=x1/y1x2/y2xn/yn，形式，3/6需要規(guī)約為1/2,3/1需要規(guī)約成3，歸約后的結(jié)果例子輸入2?1/3 例子輸出5/6

c=a*d+b*d=b*while(c!=0{z=d%c;d=c;c=}。5199410193512195711917919481

leap=year%4==0&&year%100!=||year%400==[m1,??case2：days+=28+int{intn,year,mon1,mon2,inti,j;cin>>n;for(i=1;i<=n; cin>>year>>mon1>>intmax,min; min=mon1<mon2?mon1:mon2;maxmon1mon2mon1mon2;days=0; for(j=min;j<max;{intsum; leapYear=year%4==0&&year%100!=||year%400== case1:sum=31;case2:sum=28+leapYear;break;case3:casecasecasecasecase12:sum=31;casecasecasecase11:sum=30;break;default:;}}dayssum;}if(days%7==

cout<<cout<<}return}? -----------

VINGT=V*104+I*103+N*102+G*10+TCINQ=C*103+I*103+N*10+Qif(VINGT==2*CINQ)

104（ＩＮＧＱ）*8（Ｃ）*4（Ｅ）*2（Ｖ）a從1開始，最長(zhǎng)到depthb=depth-2*c=depth-3*d=depth-4*e=depth-5*6*eadepth，則井深和各家繩子的int inta,b,c,d,for(intdepth=1;depth<=1000;for(a1adepth2a++)//a不可能{b=depth-2*a; c=depth-3*b;d=depth-4*c; e=depth-5*d;if((6*e+a)=={cout

<<“Depth:”<<depth<<endl<<"A:"<<a<<<<"B:"<<b<<<<"C:"<<c<<<<"D:"<<d<<<<"E:"<<e<<}returnreturnint

intnumberOfCases,i,num;cin>>numberOfCases;for(i=0;i<numberOfCases;{cin>>if(num%2!=0cout<<0<<""<<0<<endl;{cout<<num/4+(num%4)/2<<"";//最少cout<<(num/2)<< }}return} 例子輸 例子輸?3?35?3?3353小33小

3小

5小5小5如果有一個(gè)電池的超過了全部電池的總時(shí)如果沒有一個(gè)電池的超過全部電池的一半，總可以找到法，達(dá)到總時(shí)長(zhǎng)的一int intn;doublelast,total,longest,average;while(cin>>n)for(inti=0;i<n;for(inti=0;i<n; cin>>last; total+=last;if(last>longest)longest=}average=total/2.0;if(longest>average)cout<<fixed<<setprecision(1)<<total-longest <<endl;cout<<fixed<<setprecision(1)<<average<<}return60

5102