hdu 4282A very hard mathematic problem(枚举+2分)
hdu 4282A very hard mathematic problem(枚举+二分)
Total Submission(s): 3578 Accepted Submission(s): 1052
A very hard mathematic problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3578 Accepted Submission(s): 1052
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
题目大意很简单,解题思路也比较容易想的,先枚举z再枚举x然后二分y即可,时间用了200ms-。自己写的时候竟然因为用了库函数自带的pow函数然后有精度的损失,记得以前碰过这样的。一直以为是自己二分或者是范围枚举出错了。。。
题目地址:A very hard mathematic problem
AC代码:
#include<iostream> #include<cstring> #include<cmath> #include<string> using namespace std; long long pow1(long long a,long long p) { long long s=1; while(p) { if(p&1) s*=a; a*=a; p>>=1; } return s; } int main() { long long k; long long x,y,z; long long res,flag; while(cin>>k&&k) { res=0; for(z=2;z<=30;z++) //枚举z { long long tmp=pow(k/2.0,1.0/z); long long tmp1=pow(k*1.0,1.0/z); for(x=1;x<=tmp;x++) //枚举x { flag=0; long long l,r; l=x+1,r=tmp1; while(l<=r) //二分y { y=(l+r)>>1; long long s=pow1(x,z)+pow1(y,z)+x*y*z; if(s>k) r=y-1; else if(s<k) l=y+1; else { flag=1; break; } } if(flag) {res++;} } } cout<<res<<endl; } return 0; } /* 625 */ //187MS