HDU 4282 A very hard mathematic problem 二分
来源：http://acm.hdu.edu.cn/showproblem.php?pid=4282
题意：给出一个数n，问x^z + y^z + x*y*z = n有多少这样的x y z，其中y > x,z > 1,x,y,z都是正数。
思路：注意到z最大为31，x最大为50000，因此可以枚举x,z,二分判断y即可。
代码：
#include <iostream>#include <vector>#include <cstdio>#include <string.h>#include <vector>#include <climits>using namespace std;const int N = 50010;__int64 aa[N][35];__int64 mi(int x,int cnt){__int64 s = 1;for(int i = 1; i <= cnt; ++i)s *= x;return s;}void init(){memset(aa,-1,sizeof(aa));for(int i = 1; i < N; ++i){for(int j = 2; j <= 31; ++j){ __int64 x = mi(i,j); if(x >= INT_MAX || x < 0) continue; aa[i][j] = x;}}}int main(){//freopen("1.txt","r",stdin);init();int n;while(scanf("%d",&n) && n){int ans = 0;for(int x = 1; x < N; ++x){for(int z = 2; z <= 31; ++z){if(aa[x][z] == -1)break;int lp = 2,rp = N - 1;bool flag = false;while(lp <= rp){ int mid = (lp + rp) / 2; if(aa[mid][z] == -1){ rp = mid - 1; continue; } __int64 sum = aa[x][z] + mi(mid,z) + x*mid*z; if(sum == n ){ if(mid > x){ flag = true; break; } rp = mid - 1; } else if(sum < n){ lp = mid + 1; } else{ rp = mid - 1; }}if(flag) ans++;}}printf("%d\n",ans);}return 0;}