第九周项目六--年龄几何问题

来源：互联网发布：淘宝信用管理在哪里编辑：程序博客网时间：2024/05/18 21:06

问题及代码：

/* * Copyright (c) 2014, 烟台大学计算机学院 * All rights reserved. * 文件名称：test.cpp * 作    者：尚月 * 完成日期：2014年 10 月 27 日 * 版 本 号：v1.0 * * 问题描述：四个人的年龄成等差数列，四人年龄相加为26，相乘是880，求以他们的年龄为前四项的等差数列的前20项 * 程序输出：输出以他们的年龄为前四项的等差数列的前20项 */# include<iostream>using namespace std;int main(){    int a,n,count=0;    cout<<"等差数列为："<<endl;    for (a=1; a<=4; a++)        for (n=1; n<=6; n++)            if((4*n+6*a==26)&&(n*(n+a)*(n+a+a)*(n+a+a+a)==880))            {                ++count;                cout<<count<<":"<<n<<","<<n+a<<","<<n+a*2<<","<<n+a*3<<","<<n+a*4<<","<<n+a*5;                cout<<","<<n+a*6<<","<<n+a*7<<","<<n+a*8<<","<<n+a*9<<","<<n+a*10<<","<<n+a*11;                cout<<","<<n+a*12<<","<<n+a*13<<","<<n+a*14<<","<<n+a*15<<","<<n+a*16<<","<<n+a*17;                cout<<","<<n+a*18<<","<<n+a*19<<","<<n+a*20<<endl;            }    return 0;}

知识总结：

注意输出时逗号和结果顺序

学习心得：

做完这个程序就感觉简单多了，还需多练习。。

0 0