輸入一個正整數(shù)r,做r次下面的運算:
輸入1個正整數(shù)n,求s的前n項和,s=1/1!+2/2!+3/3!+...n/n!
3 回答
已采納

這都不會
TA貢獻(xiàn)72條經(jīng)驗 獲得超92個贊
#include<stdio.h> int?main()?{ float?sum?=?0; int?i,?j,?n,?fact; scanf("%d",?&n); for(i?=?1;?i?<=?n;?i++)?{ fact?=?1; for(j?=?i;?j?>?0;?j--)?{ fact?*=?j; } sum?+=?(1.0?*?i)/fact; } printf("%f",?sum); return?0; }

望遠(yuǎn)
TA貢獻(xiàn)1017條經(jīng)驗 獲得超1032個贊
#include<stdio.h> void?main() { int?n,m,sum; double?result; printf("請輸入你要測試的數(shù)據(jù)組數(shù):"); scanf("%d",&n); for(int?i=1;i<=n;i++) { sum=1; result=0; printf("請輸入第%d組測試數(shù)據(jù):",i); scanf("%d",&m); for(int?j=1;j<=m;j++) { sum=sum*j; result=result+1.0/sum; printf("1/%d!+",j); } printf("\b=%.4lf\n",result); } }

滿天星17
TA貢獻(xiàn)8條經(jīng)驗 獲得超1個贊
多謝!我已經(jīng)會了,程序如下:
#include<stdio.h>
int main()
{
int a,b,sum;
int repeat,i;
float s=0,t=0;
scanf("%d",&repeat);
for(a=0;a<repeat;a++)
{
scanf("%d",&b);
sum=1;
for(i=1;i<=b;i++)
{ sum*=i;
s=1.0/sum;
t+=s;
}
printf("%f\n",t);
}
return 0;
}
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