C语言，求1+1/2+1/3...+1/n

显示全部楼层 · 2019-6-6 13:12:55

# include "stdio.h"int main ( ){ int i,n; double sum; printf("请输入n:\n"); scanf("%d",&n); sum=0.0; for(i=1;i<=n;i++) {sum=sum+1.0/i; } printf("1+1/2+1/3...+1/n=%f\n",sum); return 0;} 你的错误有三点：首先是main函数的返回值应该是整型int；其次是先输出提示信息，调用printf(),然后是输入函数；最重要的是这句，sum=sum+1.0/i;把n换成i，应为n是你输入的常量！你再试一下！

千问 · 2019-6-6 13:12:55

# include "stdio.h"void main ( ){ int i,n; double sum; printf("请输入n:\n"); scanf("%d",&n);//输入和输出顺序换下sum=0.0; for(i=1;i<=n;i++) {sum=sum+1.0/i;//应该是i } printf("1+1/2+1/3...+1/n=%f",sum);}

千问 · 2019-6-6 13:12:55

# include "stdio.h"void main ( ){ int i,n; double sum=0.0; printf("请输入n:\n");scanf("%d",&n); for(i=1;i<=n;i++)sum=sum+1.0/i;printf("1+1/2+1/3...+1/n=%f\n",sum);}关键是一处：sum=sum+1.0/n;修改为sum=sum+1.0/i;

千问 · 2019-6-6 13:12:55

sum=0;少了这个语句。而且要将n改为isum=sum+1.0/i;for(i=1;i<=n;i++) {sum=sum+1.0/i; }

千问 · 2019-6-6 13:12:55

for(i=1;i<=n;i++) {sum=sum+1.0/i; }中的n改成i

千问 · 2019-6-6 13:12:55

for(i=1;i<=n;i++) {sum=sum+1.0/n; }改为for(i=1;i<=n;i++) {sum=sum+1.0/i; ////n 改成i }

千问 · 2019-6-6 13:12:55

printf("1+
虽然不大懂这应该是print吧多了个f?