hdu4282 A very hard mathematic problem-----天津网络赛
A very hard mathematic problemTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 913 Accepted Submission(s): 268
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.
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.
Output
Output the total number of solutions in a line for each test case.
Sample Input
9
53
6
0
Sample Output
1
1
0
Hint
9 = 1^2 + 2^2 + 1 * 2 * 2
53 = 2^3 + 3^3 + 2 * 3 * 3
Source
2012 ACM/ICPC Asia Regional Tianjin Online
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[cpp]
#include<iostream>
#include<cstdlib>
#include<stdio.h>
#include<math.h>
using namespace std;
#define ll __int64
ll power(ll x,int y)
{
ll temp=x;
for(int i=2;i<=y;i++)
temp*=x;
return temp;
}
int main()
{
ll n;
while(scanf("%I64d",&n)&&n)
{
int ans=0;
int temp=sqrt(n);
if(temp*temp==n) ans+=(temp-1)/2;
for(int i=3;i<31;i++)//z
{
for(ll j=1;;j++)
{
ll y=power(j,i);
if(y>=n/2) break;
for(ll k=j+1;;k++)//先枚举小的
{
ll x=power(k,i);
ll uu=x+y+i*j*k;
if(uu>n) break;
else if(uu==n){ans++;break;}
}
}
}
cout<<ans<<endl;
}
}
补充：软件开发 , C++ ,
