hdu 4282 A very hard mathematic problem(二分+枚举)
A very hard mathematic problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2581 Accepted Submission(s): 747
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
liuyiding
思路:枚举y和次数z然后二分查找是否有符合条件的x出现y*y<2^31 z增加值将是指数上涨所以时间复杂度为,log(k)*log(k)*sqrt(k)
#include<iostream> #include<cstring> using namespace std; const int mm=1000; long long yz; bool ok(long long z,long long y,long long k) { long long l=1,r=y-1,mid,xx,sum; while(l<=r) { mid=(l+r)/2; xx=1; for(int i=0;i<z;i++) xx*=mid; sum=xx+mid*yz; if(sum==k)return 1; else if(sum<k)l=mid+1; else r=mid-1; } return 0; } int main() { long long k,x,y,z,ans; while(cin>>k&&k) { ans=0;x=1;y=2;z=2; ///枚举y和次数z for(y=2;y*y<=k;y++) for(z=2;;z++) { long long yy=1; for(int i=0;i<z;i++) yy*=y; if(yy>k) break; yz=y*z; ///y,z确定,二分查找是否有符合条件的x出现 if(ok(z,y,k-yy))ans++; } cout<<ans<<"\n"; } }
The article write by nealgavin