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 = ; i <= ; i++)
         {
             if(n%i == )
             {
                 x = n/i;
                 tmp = sqrt(x);
                 if(tmp*tmp == x)
                    return x;
             }
         }
         for(i = ; i <= ; i++)
         {
             x = i*i;
             if(n%x == )
             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 = ; i <= ; i++)
         {
             if(i > n) break;  //
             if(n%i == )
             {
                 x = n/i;
                 tmp = sqrt(x);
                 if(tmp*tmp == x)
                    return x;
             }
         }
         for(i = ; i >= ; i--)  //
         {
             x = i*i;
             if(x > n) continue;  //
             if(n%x == )
             return x;
         }
     }
 };

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