标签:高精度 info main mic http ref 求导 col 数学
https://vj.z180.cn/c7965da2b8fbd28dfbce93c000a232e5?v=1578704227
题意:
高精度求和
思路:
用求导定义得出所求式子可化为对i^(1/3)求导的和*1e5,即求1/3*i^(-2/3)的和*1e15
1 #include <stdio.h> 2 #include <string.h> 3 #include <math.h> 4 5 int main() 6 { 7 int i, low, high, num; 8 double sum; 9 while(~scanf("%d %d", &low, &high)&&low&&high) 10 { 11 sum = 0.0; 12 for(i=low; i<=high; i++) 13 { 14 sum = sum + pow(i, -2.0 / 3.0) / 3.0; 15 } 16 num = 15; 17 while(sum < 1) 18 { 19 sum *= 10; 20 num++; 21 } 22 while(sum > 10) 23 { 24 sum /= 10; 25 sum--; 26 } 27 printf("%.5lfE-%03d\n", sum, num); 28 } 29 return 0; 30 }
标签:高精度 info main mic http ref 求导 col 数学
原文地址:https://www.cnblogs.com/0xiaoyu/p/12181941.html