循环结构练习参考代码

1、试验五 循环结构程序设计一、基础能力落实：1）编写一个程序，用while，该程序读取整数，直到输入0，输入终止后，程序应该报告输入的偶数（不包括0）总个数，偶数的平均值，输入的奇数总个数以及奇数的平均值。#include <stdio.h>int main() int n; double sumeven = 0.0; int ct_even = 0; double sumodd = 0.0; int ct_odd = 0; while (scanf("%d", &n) = 1 && n != 0) if (n % 2 = = 1) sum

2、odd += n; +ct_odd; else sumeven += n; +ct_even; printf("Number of evens: %d", ct_even); if (ct_even > 0) printf(" average: %f", sumeven / ct_even); printf(“n”); printf("Number of odds: %d", ct_odd); if (ct_odd > 0) printf(" average: %f", sumodd / ct_odd)

3、; printf(“n”); return 0;2）编写一个程序，将用分钟表示的时间转化成以小时和分钟表示的时间。使用#define或者const来创建一个代表60的字符常量。使用while循环来允许用户重复键入值，并且当键入一个小于等于0的时间时终止循环。要求用while语句#include <stdio.h>int main() const int minperhour = 60; int minutes, hours, mins; printf("Enter the number of minutes to convert: "); scanf("

4、;%d", &minutes); while (minutes > 0 ) hours = minutes / minperhour; mins = minutes % minperhour; printf("%d minutes = %d hours, %d minutesn", minutes, hours, mins); printf("Enter next minutes value (0 to quit): "); scanf("%d", &minutes); printf("Bye

5、n"); return 0;3）编写一个程序打印一个表，表的每一行都给出一个整数，它的平方以及它的立方，要求用户输入表的上限和下限，打印上限下限之间的数据。使用一个for循环。#include <stdio.h>int main() int lower, upper, index; int square, cube; printf("Enter starting integer: "); scanf("%d", &lower); printf("Enter ending integer: "); scan

6、f("%d", &upper); printf("%5s %10s %15sn", "num", "square", "cube"); for (index = lower; index <= upper; index+) square = index * index; cube = index * square; printf("%5d %10d %15dn", index, square, cube); return 0;4) 编写一个程序，输出100200

7、的全部素数。#include <stdio.h>#include <math.h>int main()int m,i,k,n=0;for(m=101;m<=200;m=m+2)k=sqrt(m);for(i=2;i<=k;i+)if(m%i = = 0)break;if(i= =k+1)printf(“%d ”,m);n=n+1;if(n%10 = = 0)printf(“n”);printf(“n”);5）编写一个程序打印出下列图形：*#include <stdio.h>int main()int i,j;for(i=0;i<4;i+)f

8、or(j=0;j<=i ;j+)printf(“*”);printf(“n”);6）编写一个程序，输出100999之间的所有各位数之和为10的数。 #include <stdio.h>int main()int i,k=0;for(i=100;i<1000;i+)if( ( i / 100 + ( i /10 ) % 10 + i%10 ) = 10) printf(“%d “, i); k+;if(k= =10)printf(“n”); k = 0; printf(“n”);return 0;二、进阶能力落实：1) 编写一个程序，输出如下图案： * *#include

9、 <stdio.h>int main()int i,j;for (i=0;i<=3;i+)for(j=0;j<=2-i;j+)printf(“ ”);for(j=0;j<=2*i;j+)printf(“*”);printf(“n”);for(i=0;i<=2;i+)for(j=0;j<=i;j+)printf(“ ”);for(j=0;j<=4-2*i;j+) printf(“*”);printf(“ n”);return 0;2)编写一个程序，找出1000以内所有的完数，所谓完数就是一个数恰好等于它的因子之和，例如6=1+2+3，6就是一个完数

10、。#include <stdio.h>int main()int m,s,i;for(m=1;m<=1000;m+)s=0;for(i=1;i<m;i+) if(m%i= =0) s=s+i; if(s = = m)printf(“%d ”,m);printf(“n”);return 0;6.3#include <stdio.h>int main() int a,n,i; int sum=0; printf("please input a integer(1-9): "); scanf("%d", &a); p

11、rintf("please input a integer(1-6): "); scanf("%d", &n); for(i=0;i<n;i+) sum += a; printf("%d ", a); a = 10*a + a%10; printf(": %dn", sum); return 0;6.4#include <stdio.h>#define N 5 int main() int i,count=1,sum=0; for(i=1;i<=N;i+) count = count * i; sum += count; printf("%d! + ", i); printf("%d! = %dn", i, sum); return 0;6.8#include <stdio.h>#define N 5int main() int i; float numerator,denominator,temp,sum=0; numerator = 2; denominator = 1; for(i=0;i<N;i+) sum += numerator /