用C语言如何编写1-1/2+1/3-1/4+…+(-1)(n+1次方)/n c语言用函数编写:1-1/2+1/3-1/4.....+((...

\u7528C\u8bed\u8a00\u5982\u4f55\u7f16\u51991-1/2+1/3-1/4+\u2026+(-1)(n+1\u6b21\u65b9)/n

\u7ed9\u4f60\u4e2a\u7528\u9012\u5f52\u65b9\u6cd5\u7684\uff0c\u8fd9\u79cd\u9898\u76ee\u4e00\u822c\u662f\u7528\u9012\u5f52\u7b97\u6cd5\u505a\u7684
#include
#include
long double fun(int n);
main()
{
int n;
printf("\u8bf7\u8f93\u5165n\u7684\u503c\uff1a\n");
scanf("%d",&n);
printf("\u8868\u8fbe\u5f0f\u7684\u503c\u4e3a:%lf\n",fun(n));
}
long double fun(int n)
{
long double result;
if(n == 1)
{
return 1;
}
else
{
result = (long double)pow(-1,n+1)/n;
return result+fun(n-1);
}
}

#includedouble fun(int n){double s=0; for(;n;) if(n%2)s+=1.0/n--; else s-=1.0/n--; return s;}int main(){int n; scanf("%d",&n); printf("%g\n",fun(n)); return 0;}

这样想(-1)的N次方取决于N的奇偶性,所以只要判断N是偶还是奇就可以了#include<stdio.h>int ncf(int n) { if (n==(n/2)*2) return(1); else return(-1);}//判断奇偶,如果是奇返回-1 否则返回1int main(){ int n,i; float f=0; scanf("%d",&n); for (i=1;i<=n;i++) f+=ncf(i+1)/(float)i;//计算1-1/2+1/3-1/4+…+(-1)(n+1次方)/n printf("%f",f);}

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