ACM HD C++ 1012此文为了揭露c++中精确到结果小数位数的方法

来源：互联网发布：淘宝搭配套餐编辑：程序博客网时间：2024/05/16 05:49

u Calculate e

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 42504 Accepted Submission(s): 19426

Problem Description

A simple mathematical formula for e is

where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.

Output

Output the approximations of e generated by the above formula for the values of n from 0 to 9. The beginning of your output should appear similar to that shown below.

Sample Output

n e- -----------0 11 22 2.53 2.6666666674 2.708333333

Source

Greater New York 2000

Recommend

JGShining | We have carefully selected several similar problems for you: 1013 1021 1017 1020 1014

#include<iostream>#include<iomanip> //精确到小数点后几位需要的头文件#define max 13using namespace std;int factorial(int n){int k=1;if (n == 0) return 1;for (int i = 1; i <= n; i++)k = k*i;return k;}int main(){double c[max];memset(c, 0, sizeof(int)*max);cout << "n e" << endl;cout << "- -----------" << endl;c[0] = factorial(0);for (int i = 1; i <= 10; i++){c[i] = c[i - 1] + 1.0 / factorial(i);if (i <= 2){cout << i - 1 << " " << fixed << setprecision(0) << c[i - 1] << endl;}else if (i == 3)cout << i - 1 << " " << fixed << setprecision(1) << c[i - 1] << endl;else{cout << i-1 << " " << fixed << setprecision(9) << c[i-1] << endl; // 处理数据的格式}}return 0;}</iomanip></iostream>

0 0