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第九周项目六--年龄几何问题

2014-10-27 16:33 232 查看
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。。
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