hdu4282 A very hard mathematic problem-天津市网络赛
hdu4282 A very hard mathematic problem-----天津网络赛
Total Submission(s): 913 Accepted Submission(s): 268
A very hard mathematic problem
Time 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.
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
2012 ACM/ICPC Asia Regional Tianjin Online
Recommend
liuyiding
#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; } }