include<stdio.h>int main(){ double sum=0; int i,n; scanf("%d",&n); for(i=1;i<=n;i++) { sum=sum+1/(double)(4*i-3)-1/(double)(4*i-1);//要转换为浮点型的 } printf("%.6lf\n",sum*4); return 0;} ...
int main(){double pi=1,t=0;for(;t<1.9999999;){t=sqrt(2+t);pi*=2/t;} printf("%lf\n",pi*2);return 0;}
可以这样，include <stdio.h>int main(){double d1 = 1234.567;double d2 = d1 - (int)d1; // 求出小数部分int i = d2 * 1000; // 取出小数后3位if (i%10 >=5 ){i -= i%10;i += 10; // 四舍五入}d1 = (int)d1 + (double)i / 1000;printf("d1 = %lf \n",...