简单C语言程序，while循环写一个程序求数列1+1/2+1/3+……1/n的和并将结果输出 如何在C编程语言用do-while语句编写程序计算e≈1+1...
\u7f16\u7a0b\uff0c\u7528while\u8bed\u53e5\uff0c\u8ba1\u7b97\u5e76\u8f93\u51fa1+1/2+\u2026+1/n\u7684\u503c#include
int n,i;
float sum=0.0;
int main()
{
scanf("%d",&n);
for(i=1;i<=n;i++)
sum+=1.0/i;
printf("%f\n",sum);
getch();
return 0;
}
\u4f60\u597d
do-while\u8bed\u53e5\u5b9e\u73b0\u5982\u4e0b\uff1a
#include
int f(int n){/*\u81ea\u5b9a\u4e49\u4e00\u4e2a\u6c42\u9636\u4e58\u7684\u51fd\u6570*/
if(n==1||n==0)
return 1;
return n*f(n-1);
}
int main(){
double index,s=0;
int n,i;
printf("\u8f93\u5165\u4e00\u4e2a\u6b63\u6574\u6570n(n<34):");
scanf("%d",&n);
i=0;
do{
index=1/(1.0*f(i));
s+=index;
i++;
}while(i<=n);
if(n==0)
printf("e\u22481");
else if(n==1)
printf("e\u22482");
else
printf("e\u22481+1/1!+\u2026\u2026+1/%d!=%f\n",n,s);
return 0;
}
\u5e0c\u671b\u80fd\u5e2e\u52a9\u4f60\u54c8
void F1(int n)
{
double result = 0;
for(double i=1;i<=n;i++)
result += 1/i;
for(int i=1;i<n;i++)
{
printf("1/%d + ", i);
}
printf("1/%d = %f
", n, result);
}
#include <stdio.h>
int main(void)
{
int i, n;
float sum = 0;
scanf("%d",&n);
for(i = 1; i <= n; i++)
sum = sum + 1/i;
printf("%f", sum);
printf("\n");
return 0;
}
不知道有没错误，大概就这样。
书上这样的例子挺多的吧。
#include <stdio.h>
int main(void)
{
int i,n;
double sum=0.0;
printf("please input the munber:\n");
scanf("%d",&n);
i = 1;
while (i<=n)
{
sum+=1.0/i;
i++;
}
printf("sum=%lf\n",sum);
return 0;
}
其实感觉还是用for比较直观，方便，希望能帮到你。
int main(void)
{
int i = 1, n , sum=0;
scanf("%d",&n);
while(i > n)
{
i +=1;
sum += 1/ i;
}
printf("%d",sum);
}
#include <stdio.h>
int main()
{
int i,n;
float sum=0.0;
scanf("%d",&n);
for(i=1;i<=n;i++)
sum+=1.0/i;
printf("sum=%f\n",sum);
return 0;
}
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