根据下面关系式,求圆周率的值,直到最后一项的值小于给定阈值。
2π=1+31+3×52!+3×5×73!+⋯+3×5×7×⋯×(2n+1)n!+⋯
输入格式:
输出格式:
输入样例:
0.01
输出样例:
3.132157
思路:得使用double类型
#include<stdio.h>int main(){int i=0;double fz,fm,fs;double yz,sum;scanf("%lf", &yz);fm = fz = 1;sum = fs = fz/fm;while(fs>=yz){i++;fz = fz * i;fm = fm * ( 2*i+1 );fs = fz/fm;sum = sum + fs;}printf("%.6lf", 2*sum);return 0;}
