1012 u Calculate e

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u Calculate e

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45362 Accepted Submission(s): 20825

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

代码

#include <iostream>#include <string>using namespace std;int main(){cout << "n e" << endl;cout << "- -----------" << endl;cout << "0 1" << endl;cout << "1 2" << endl;cout << "2 2.5" << endl;double sum = 2.5;int a = 1 * 2;cout.setf(ios::fixed);cout.precision(9);for (int i = 3; i < 10; ++i){a *= i;sum += (double)1 / a;cout << i << " " << sum << endl;}return 0;}

本题就是按格式循环输出结果，前面三行的格式与后面的格式不一致，直接输出；之后3~9用for循环计算并输出

cout.setf(ios::fixed);表示定点输出，即不会表示成科学计数
scientific则是表示以科学记数法表示

cout.precision(9);表示输出的小数点位数

总之，很明显cout的格式输出没有printf来的好用，直接用printf就可以了

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