第九周项目六穷举法解决组合问题（3）年龄几何

/*
* Copyright (c) 2014, 烟台大学计算机学院
* All rights reserved.
* 文件名称：test.cpp
* 作    者： 郝俊宇
* 完成日期：2014年 10 月 25日
* 版 本 号：v1.0
*
* 问题描述：张三，李四，王五，刘六的年龄成一等差数列，他们四人的年龄相加是26，相乘是880，求以他们的年龄为前4项的等差数列的前20项
* 程序输出：一个数代表前20项的和
*/
#include <iostream>

using namespace std;

int main()
{
int n,a;
cout<<"数列为：";
for(n=1; n<=6; n++)
for(a=1; a<=4; a++)
{
if((4*n+6*a==26)&&(n*(n+a)*(n+a+a)*(n+a+a+a)==880))
{

cout<< n<<","<<n+a<<","<<n+2*a<<","<<n+3*a<<","<<n+4*a<<","<<n+5*a<<"，"<<n+6*a<<","<<n+7*a<<","<<n+8*a<<","<<n+9*a<<","<<
n+10*a<<","<<n+11*a<<","<<n+12*a<<","<<n+13*a<<","<<n+14*a<<","<<n+15*a<<","<<n+16*a<<","<<n+17*a<<"，"<<n+18*a<<","<<
n+19*a<<endl;
}
}
return 0;
}