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 0
Hint
9 = 1^2 + 2^2 + 1 * 2 * 2
53 = 2^3 + 3^3 + 2 * 3 * 3
题意:给你一个整数k,问你x^z+y^z+z*x*y=k的解有多少种,其中y>x>0,k<2^31;
1.因为x最小为1,且y>x,所以y最小为2;
2.因为x,y>0,且k<2^31,所以z<=30;
3.当z=2时,x^2+y^2+2*x*y=(x+y)^2=k;也就是说当z=2时式子有解时,k是一个完全平方数,也就是求y>x时(x+y)^2=k的解法有多少种;
#include<stdio.h> #include<string.h> #include<algorithm> #include<math.h> using namespace std; long long pow(long long x,int y) { long long ans=1; for(int i=1;i<=y;i++) ans*=x; return ans; } int main() { int k; while(~scanf("%d",&k)) { if(k==0) break; long long sum=0; long long tx,ty; long long t=(int)sqrt(k*1.0);//当z为2时; if(t*t==k)//判断是不是k是不是完全平方数; sum+=(t-1)/2; for(int z=3;z<31;z++)//z最小从3开始 { for(long long x=1; ;x++) { tx=pow(x,z); if(tx>=k/2) break; for(long long y=x+1; ;y++) { ty=pow(y,z); if(tx+ty+y*x*z>k)//大于k的时候结束; break; else if(tx+ty+x*y*z==k) { sum++; break;//记得break; } } } } printf("%lld\n",sum); } return 0; }