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

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdlib>
 4 #include <cmath>
 5 #include <cstdio>
 6 #include <vector>
 7 #include <algorithm>
 8 #define LL long long
 9 using namespace std;
10 
11 class SquareDivisor
12 {
13 public:
14     long long biggest(long long n)
15     {
16         long long i, x, tmp;
17         for(i = 1; i <= 1000000; i++)
18         {
19             if(n%i == 0)
20             {
21                 x = n/i;
22                 tmp = sqrt(x);
23                 if(tmp*tmp == x)
24                    return x;
25             }
26         }
27         for(i = 1; i <= 1000000; i++)
28         {
29             x = i*i;
30             if(n%x == 0)
31             return x;
32         }
33     }
34 };
35 
36 int main()
37 {
38     long long n, ans;
39     SquareDivisor xx;
40     while(cin>>n)
41     {
42         ans = xx.biggest(n);
43         cout<<ans<<endl;
44     }
45 }

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。

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdlib>
 4 #include <cmath>
 5 #include <cstdio>
 6 #include <vector>
 7 #include <algorithm>
 8 #define LL long long
 9 using namespace std;
10 
11 class SquareDivisor
12 {
13 public:
14     long long biggest(long long n)
15     {
16         long long i, x, tmp;
17         for(i = 1; i <= 1000000; i++)
18         {
19             if(i > n) break;  //
20             if(n%i == 0)
21             {
22                 x = n/i;
23                 tmp = sqrt(x);
24                 if(tmp*tmp == x)
25                    return x;
26             }
27         }
28         for(i = 1000000; i >= 1; i--)  //
29         {
30             x = i*i;
31             if(x > n) continue;  //
32             if(n%x == 0)
33             return x;
34         }
35     }
36 };
37 
38 int main()
39 {
40     long long n, ans;
41     SquareDivisor xx;
42     while(cin>>n)
43     {
44         ans = xx.biggest(n);
45         cout<<ans<<endl;
46     }
47 }

 

posted @ 2014-07-28 16:51  水门  阅读(135)  评论(0)  编辑  收藏  举报