tc 2014 college tour 250 500

题意：

You are given a long long n. Return the largest divisor of n that is a perfect square. That is, the correct return value is x if and only if:

1. x divides n
2. There is an integer y such that x = y*y.
3. x is the largest integer that satisfies conditions 1 and 2.

求最大的y*y 使N MOD (y*y)=0;

x = y*y 返回x的值。

分析：

假设 x*t = n;

当t<=10^6时，枚举t,同时查看是否符合y*y == x的条件。

当t>10^6时，即x<=10^12, y<=10^6, 此时枚举y,

所以两个10^6的循环就可以了。

500:  错误代码

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <cstdio>
#include <vector>
#include <algorithm>
#define LL long long
using namespace std;

class SquareDivisor
{
public:
    long long biggest(long long n)
    {
        long long i, x, tmp;
        for(i = 1; i <= 1000000; i++)
        {
            if(n%i == 0)
            {
                x = n/i;
                tmp = sqrt(x);
                if(tmp*tmp == x)
                   return x;
            }
        }
        for(i = 1; i <= 1000000; i++)
        {
            x = i*i;
            if(n%x == 0)
            return x;
        }
    }
};

int main()
{
    long long n, ans;
    SquareDivisor xx;
    while(cin>>n)
    {
        ans = xx.biggest(n);
        cout<<ans<<endl;
    }
}

先保存一下，不知道为什么错了

知道错哪了：首先没有判断 x>n的时候停止， 而且第二个循环的时候从前向后 return 的不是最大值。

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <cstdio>
#include <vector>
#include <algorithm>
#define LL long long
using namespace std;

class SquareDivisor
{
public:
    long long biggest(long long n)
    {
        long long i, x, tmp;
        for(i = 1; i <= 1000000; i++)
        {
            if(i > n) break;  //
            if(n%i == 0)
            {
                x = n/i;
                tmp = sqrt(x);
                if(tmp*tmp == x)
                   return x;
            }
        }
        for(i = 1000000; i >= 1; i--)  //
        {
            x = i*i;
            if(x > n) continue;  //
            if(n%x == 0)
            return x;
        }
    }
};

int main()
{
    long long n, ans;
    SquareDivisor xx;
    while(cin>>n)
    {
        ans = xx.biggest(n);
        cout<<ans<<endl;
    }
}