如题所述
#include<stdio.h>
void main()
{
int n=0;
printf("请输入数字：\n");
scanf("%d",&n);
double sum=0;
for(int i=1;i<=n;i++){
sum+=1*1.0/i;
}
printf("1+1/2+1/3+....+1/n = %lf\n",sum);
}
void main()
{
int n=0;
printf("请输入数字：\n");
scanf("%d",&n);
double sum=0;
for(int i=1;i<=n;i++){
sum+=1*1.0/i;
}
printf("1+1/2+1/3+....+1/n = %lf\n",sum);
}
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第1个回答 2015-06-04
#include <>
#include <>
int main() {
int i, n;
double sum = 0.0f;
printf("Input N:");
scanf("%d", &n);
if (n<1) {
printf("The input N is invalid.\n");
exit(0);
}
for (i=1;i<=n;i++) {
sum += 1.0/(double)i;
}
printf("Result:%.9f\n", sum);
return 0;
}本回答被提问者采纳
#include <>
int main() {
int i, n;
double sum = 0.0f;
printf("Input N:");
scanf("%d", &n);
if (n<1) {
printf("The input N is invalid.\n");
exit(0);
}
for (i=1;i<=n;i++) {
sum += 1.0/(double)i;
}
printf("Result:%.9f\n", sum);
return 0;
}本回答被提问者采纳
