P39-习题2-4

/*输入两个正整数m,n,求(i^2+1/i),从m到n的和*/
#include<stdio.h> int main(void) { int m,n,i; double product,result; /*result为单个元素的解,product为所有解的和*/
printf(
"Enter m:"); scanf("%d",&m); printf("Enter n:"); scanf("%d",&n); product=0; /*从m到n依次循环求解*/
for(i=m;i<=n;i++){ result=(i*i)+(1/i); printf("%d",result); /*将单个元素所求解累加起来*/
product
=product+result; } printf("product=%.2f ",product); return 0; }
原文地址:https://www.cnblogs.com/xym0914/p/3373297.html