HDU 4282 A very hard mathematic problem
时间：2014-05-11 12:57:48
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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4287
Problem Description
Haoren is very good at solving mathematic problems. Today he is working a problem like this:
Find three positive integers X, Y and Z (X < Y, Z > 1) that holds
X^Z + Y^Z + XYZ = K
where K is another given integer.
Here the operator “^” means power, e.g., 2^3 = 2 * 2 * 2.
Finding a solution is quite easy to Haoren. Now he wants to challenge more: What’s the total number of different solutions?
Surprisingly, he is unable to solve this one. It seems that it’s really a very hard mathematic problem.
Now, it’s your turn.
Find three positive integers X, Y and Z (X < Y, Z > 1) that holds
X^Z + Y^Z + XYZ = K
where K is another given integer.
Here the operator “^” means power, e.g., 2^3 = 2 * 2 * 2.
Finding a solution is quite easy to Haoren. Now he wants to challenge more: What’s the total number of different solutions?
Surprisingly, he is unable to solve this one. It seems that it’s really a very hard mathematic problem.
Now, it’s your turn.
Input
There are multiple test cases.
For each case, there is only one integer K (0 < K < 2^31) in a line.
K = 0 implies the end of input.
For each case, there is only one integer K (0 < K < 2^31) in a line.
K = 0 implies the end of input.
Output
Output the total number of solutions in a line for each test case.
Sample Input
9 53 6 0
Sample Output
1 1 0Hint9 = 1^2 + 2^2 + 1 * 2 * 2 53 = 2^3 + 3^3 + 2 * 3 * 3
Source
Recommend
题意：
给你一个式子x^z + y^z + x*y*z = k，k 为给定的某个int 范围内的数字。
求共有多少组关于x,y,z 的解。（0< x < y,z > 1）
解题思路：
显然当z 越大的时候x,y 的值越小。
由于y 最小等于2，所以有2^z < k，又k < 2^31，所以有z < 31。
1、首先考虑当z=2 的时候，式子左边可以化为(x+y)^2 = k 的形式。
所以当k 是完全平方数的时候，(x,y) 才有可能有解。
假如m^2 = k ，问题就相当于求x+y = m (x < y) 的解的组数。
容易得出ans=(m-1)/2;
2、然后考虑当z>=3 的时候。
枚举就好！
代码如下：
#include<cstdio>
#include<cmath>
#define LL __int64
LL POW(LL x , int z)
{
LL ans = 1;
for(int i = 1 ; i <= z ;i++)
{
ans*=x;
}
return ans;
}
int main()
{
LL x,y,tx,ty,ans;
int z,k;
while(~scanf("%d",&k) && k)
{
ans = 0;
LL tmp = (int)sqrt(k*1.0);
if(tmp *tmp == k)//若n是一个完全平方数
ans+=(tmp-1)/2;
for( z = 3 ; z < 31 ; z++)//z最大为30，开始枚举
{
for(LL x = 1 ; ; x++)
{
tx = POW(x,z);
if(tx >= k/2)
break;
for(LL y = x+1; ; y++)
{
ty = POW(y,z);
if(tx+ty+x*y*z > k)
break;
else if(tx+ty+x*y*z == k)
{
ans++;
break;
}
}
}
}
printf("%I64d\n",ans);
}
return 0;
}
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