zoj 1239 Hanoi Tower Troubles Again!

Hanoi Tower Troubles Again!

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Time Limit: 2 Seconds Memory Limit: 65536 KB

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People stopped moving discs from peg to peg after they know the number of steps needed to complete the entire task. But on the other hand, they didn't not stopped thinking about similar puzzles with the Hanoi Tower. Mr.S invented a little game on it. The game consists of N pegs and a LOT of balls. The balls are numbered 1,2,3... The balls look ordinary, but they are actually magic. If the sum of the numbers on two balls is NOT a square number, they will push each other with a great force when they're too closed, so they can NEVER be put together touching each other.

 

The player should place one ball on the top of a peg at a time. He should first try ball 1, then ball 2, then ball 3... If he fails to do so, the game ends. Help the player to place as many balls as possible. You may take a look at the picture above, since it shows us a best result for 4 pegs.

Input

The first line of the input contains a single integer T, indicating the number of test cases. (1<=T<=50) Each test case contains a single integer N(1<=N<=50), indicating the number of pegs available.

Output

For each test case in the input print a line containing an integer indicating the maximal number of balls that can be placed. Print -1 if an infinite number of balls can be placed.

Sample Input

2
4
25

Sample Output

11
337

 

不是很难的题，看懂题意就好，只需要记录每根柱子的第一个球是多少就好。

 

题意：给定柱子输N，按编号从小到大放球，要求，如果该球不在最底数，则该球和它下面一个球的编号之和必须是完全平方数。

给定柱子数N，最多能放多少球上去。如果无限个，就输出-1（我没看出什么时候输出-1）

 

附上代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

int a[55],n,m;

bool xx(int x)  //x是当前要放的球的编号
{
    int s,w;
    for(int i=1; i<=n; i++)   //一共有这么多的柱子
    {
        s=a[i]+x;
        w=(int)sqrt(s)*(int)sqrt(s);
        if(!a[i]||s==w)  //当柱子为空或者此时和为完全平方数时，放入这个球
        {
            a[i]=x;
            m++;
            xx(x+1);
            return true;
        }
    }
    return false;
}

int main()
{
    int T,i,j;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        memset(a,0,sizeof(a)); //a数组存储最上面的球的编号
        m=0;
        if(xx(1)) //从编号1开始循环
            printf("%d\n",m);
        else
            printf("-1\n");
    }
    return 0;
}