设计程序sy3-1.c,计算并输出n(包括n)以内所有能被5或9整除的自然数的倒数之和？

#include <stdio.h>

int main() {

int i,n

double sum

printf("n = ")

while(scanf("%d",&n) == 1) {

sum = 0.0

for(i = 5i <= n++i) {

if((i%5 == 0) &&(i%9 == 0)) {

sum += 1.0/i

printf("%d ",i)

}

}

if(sum >0)

printf("\n1 － %d之间满足条件的自然数倒数之和为:%lf\n",n,sum)

printf("\nn = ")

}

printf("\n")

return 0

}

#include <stdio.h>

int main(int argc,char **argv)

{

double res = 0

int i = 0

double jishu = 1.0

for(i = 1i <= 10000i++)

{

res += jishu/i

}

printf("1到10000的倒数和为%f",res)

}

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