HDU - 1012ACM

这题就更简单了。

不说了~ 上个代码就好了。

////  main.cpp//  uCalculatee////  Created by Alps on 14/12/25.//  Copyright (c) 2014年 chen. All rights reserved.//#include <iostream>using namespace std;double jiecheng(double b){    double a = 1;    while (b > 0) {        a = a * b;        b -= 1;    }    return a;}int main(int argc, const char * argv[]) {    double a = 1,b = 1;    int i;    printf("n e\n");    printf("- -----------\n");    printf("0 1\n");    for (i = 1; i <= 1; i++) {        a = a + (1/jiecheng(b));        printf("%d %.0f\n",i,a);        b += 1;    }    for (; i <= 2; i++) {        a = a + (1/jiecheng(b));        printf("%d %.1f\n",i,a);        b += 1;    }    for (; i <= 9; i++) {        a = a + (1/jiecheng(b));        printf("%d %.9f\n",i,a);        b += 1;    }    return 0;}