c語言程序答案

1、#include <stdio.h> main() (const double pi = 3.14159;double r;double circum;double area;printf("Input r:"); scanf("%lf", &r); circum = 2 * pi * r; area = pi * r * r; printf("printf WITHOUT width or precision specifications:n"); printf("circumference = %f,

2、 area = %fn", circum, area);printf("printf WITH width and precision specifications:n"); printf("circumference = %7.2f, area = %7.2fn", circum, area); 使用const常量定義圓周率 pi (取值為3.14159),編程從鍵盤輸入圓的半徑r,計(jì)算并輸出圓的周長(zhǎng)和面積.輸出的數(shù)據(jù)保存兩位小數(shù)點(diǎn).*輸入格式要求："lf"提示信息："Input r:" *輸出格式要求

3、："printf WITHOUT width or precision specifications:n" "circumfer ence = %f, area = %fn" "printf WITH width and precision specifications:n" "circumfere nce = %7.2f, area = %7.2fn"程序運(yùn)行例如如下： Input r:5.3 printf WITHOUT width or precision specifications: circumfere

4、nce = 33.300854, area = 88.247263 printf WITH wid th and precision specifications: circumference = 33.30, area = 88.25#include<stdio.h> main()(int a,b,c;double p,q,x1,x2;printf("Please enter the coefficients a,b,c:");scanf("%f,%f,%f",&a,&b,&c);p=-b/a;q=(bA2-4ac)

5、A0.5/(2a);x1=p-q;x2=p+q;#include<stdio.h>Main()(int a,b,c;double p,q,x1,x2;printf("Please enter the coefficients a,b,c:");scanf("%f,%f,%f",&a,&b,&c);p=-b/a;q=(bA2-4ac)A0.5/(2a);x1=p-q;x2=p+q;)計(jì)算輸入的兩個(gè)整數(shù)的和.*輸入格式要求："d提示信息：請(qǐng)輸入第一個(gè)整數(shù)：""請(qǐng)輸入第二個(gè)整數(shù)："*輸

6、出格式要求："d + %d = %dn"程序運(yùn)行例如如下： 請(qǐng)輸入第一 個(gè)整數(shù)：34請(qǐng)輸入第二個(gè)整數(shù)： 56 34 + 56 = 90#include <stdio.h> int main()(int n1;定義第一個(gè)整型變量n1int n2;定義第一個(gè)整型變量n2printf("請(qǐng)輸入第一個(gè)整數(shù)：")；務(wù)必養(yǎng)成好習(xí)慣：任何輸入前先給出輸入提示!scanf("%d", &n1);接收第一個(gè)整數(shù) n1printf("請(qǐng)輸入第二個(gè)整數(shù)：")；務(wù)必養(yǎng)成好習(xí)慣：任何輸入前先給出輸入提示!scanf(&q

7、uot;%d", &n2);接收第二個(gè)整數(shù) n2printf("%d + %d = %dn", n1, n2, n1 + n2);/*每一個(gè)格式限制符 d限制后面的一項(xiàng),第一個(gè) d限制n1 ,第二個(gè)d限制n2第三個(gè)限制n1+n2 , %用于限制輸出項(xiàng)以十進(jìn)制有符號(hào)整數(shù)的形式輸出,其它的格式限制字符串中的字符,如+,=等原樣輸出,n為換行符*/return 0; )學(xué)校有近千名學(xué)生(大于等于900,小于1000),在操場(chǎng)上排隊(duì),5人一行余2人,7人一*輸出格式要求："there a行余3人,3人一行余1人,編寫一個(gè)程序求該校的學(xué)生人數(shù).re %d

8、students in the groundn"#include <stdio.h> int main()int n;for | (n = 900; n < 1000; n+)"|if (n % 5 = 2 && n % 7 = 3&& n % 3 = 1)printf("there are %d students in the groundn", n);)(假設(shè)對(duì)折一弓厚度為 0.1mm的紙,請(qǐng)問要對(duì)折多少次才能使紙的厚度從地球到達(dá)月球(假 設(shè)地球到月球的距離為30萬km)?*輸出格式要求："

9、%d"#include <stdio.h>#include <math.h>int main() printf("%d", (int)ceil(log10(3.0) + 12) / log10(2.0);return 0;請(qǐng)使用short int型的變量計(jì)算兩個(gè) 7位十進(jìn)制整數(shù)的和,要求使用盡可能少的變量.* *輸入格式要求："a = %3hd%4hd, b = %3hd%4hd"* *輸出格式要求："a + b = %d"程序運(yùn)行例如如下：a = 1234567,b = 7654321輸出a + b

10、 = 8888888#include <stdio.h>int main()short int d, m, c, n;scanf("a = %3hd%4hd, b = %3hd%4hd", &d, &m, &c, &n);printf("a + b = %d", (d + c) * 10000 + m + n);return 0;華氏和攝氏溫度的轉(zhuǎn)換公式為：C = 5/9 x (F-32)其中,C表示攝氏溫度,F表示華氏溫度.要求：華氏0 丁300 下,每隔20 丁輸出一個(gè)華氏溫度對(duì)應(yīng)的攝氏溫度值.* *輸入提

11、示信息要求：無* *輸入數(shù)據(jù)格式要求：無* *輸出數(shù)據(jù)格式要求："4.0f%10.1fn"(第一個(gè)浮點(diǎn)數(shù)為華氏溫度,第二個(gè)浮點(diǎn)數(shù)為攝氏溫度)#include <stdio.h>int main()int upper, step;float f = 0, c;upper = 300;step = 20;while (f <= upper)c = 5.0 / 9 * (f - 32);printf("%4.0f%10.1fn", f, c);f = f + step;)return 0;編程輸出以下形式的乘法九九表:1*1= 1 2*1=

12、2 2*2= 4 3*1= 3 3*2= 6 3*3= 9 4*1 =4 4*2= 8 4*3=12 4*4=16 5*1= 5 5*2=10 5*3=15 5*4=20 5*5=25 6*1= 6 6*2=12 6* 3=18 6*4=24 6*5=30 6*6=36 7*1= 7 7*2=14 7*3=21 7*4=28 7*5=35 7*6=42 7*7=498*1= 8 8*2=16 8*3=24 8*4=32 8*5=40 8*6=48 8*7=56 8*8=64 9*1= 9 9*2=18 9*3=27 9*4=36 9*5=45 9*6=54 9*7=63 9*8=72 9*9=

13、81 *輸出格式要求："%1d*%1d=%2d #include <stdio.h> main() int i,j;for(i=1;i<=9;i+)for(j=1;j<=i;j+)-1printf("%d*%d=%2d ",i,j,i*j);printf("n");小明今年12歲,他母親比他大 24歲.編寫一個(gè)程序計(jì)算小明的母親在幾年后比小明的年 齡大一倍,那時(shí)他們兩人的年齡各是多少*輸出格式要求："year=%dn" "mingAge=%dn" "motherAge=%

14、dn" 程序運(yùn)行例如如下：year=12mingAge=24motherAge=48#include <stdio.h> int main() int mingAge, motherAge, year;mingAge = 12;motherAge = mingAge + 24;year = 1;while (motherAge + year) != 2 * (mingAge + year) year = year + 1;)printf("year=%dn", year);printf("mingAge=%dn", mingAge

15、+ year);printf("motherAge=%dn", motherAge + year);return 0;)用二維數(shù)組作為函數(shù)的參數(shù),實(shí)現(xiàn)并輸出矩陣的轉(zhuǎn)置和轉(zhuǎn)置后矩陣副對(duì)角線元素之和.按如下函數(shù)原型編程計(jì)算. 其中n的值由用戶從鍵盤輸入. 使用ProduceMatrix函數(shù)生成n階方 陣,以start值作為矩陣的第一個(gè)元素 all ,每行下一個(gè)元素比該行上一個(gè)元素大1,每行第一個(gè)元素比上一行最后一個(gè)元素大1.例如n=3,start=3時(shí)ProduceMatrix生成的矩陣為：3456789 1011m和n的取值范圍為1,30.函數(shù)原型為void Transpos

16、e(int a口N,int n);void ProduceMatrix(int a口N,int n,int start);void PrintMatrix(int atN, int n);int SumCounterDiagonal(int aN,int n); / 求矩陣副對(duì)角線元素之和 輸入提示信息："Input n:n" "Input %d*%d matrixs start element:n 輸入格式："d"輸出提示信息和格式："The transposed matrix is:n""The sum of

17、the matrix-counter-diagonal is:%dn"輸出格式(輸完矩陣的一行要換行)："4d注意除了 PrintMatrix外的輸入輸出均在 main函數(shù)里面#include<stdio.h>#define N 30void Transpose(int a口N, int n);void ProduceMatrix(int a口N, int n, int start);void PrintMatrix(int a口N, int n);int SumCounterDiagonal(int aN, int n);int main() int n ,

18、sum = 0, start = 0;int aNN = 0;/1printf("Input n:n");scanf("%d", &n);printf("Input %d*%d matrixs start element:n", n, n);scanf("%d", &start);/1ProduceMatrix(a, n, start);Transpose(a, n);/1printf("The transposed matrix is:n");PrintMatrix(a, n);/1sum = SumCounterDiagonal(a, n);printf("The sum of the matrix-counter-diagonal is:%dn", sum); return 0; void Produc