HDU 4282 A very hard mathematic problem 二分题目

http://acm.hdu.edu.cn/showproblem.php?pid=4282

题解：http://www.cnblogs.com/E-star/archive/2012/09/11/2680992.html

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>

#define CL(a,num) memset((a),(num),sizeof(a))
#define iabs(x)  ((x) > 0 ? (x) : -(x))
#define Min(a,b) (a) > (b)? (b):(a)
#define Max(a,b) (a) > (b)? (a):(b)

#define ll __int64

#define MOD 100000007
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define test puts("<------------------->")
#define maxn 100007
#define M 100007
#define N 107
using namespace std;
//freopen("din.txt","r",stdin);

ll inf = 2147483648ll;

ll Pow(int a,int b){
    ll sum = 1;
    for (int i = 1; i <= b; ++i) sum *= a;
    return sum;
}
bool Bsearch(int L,int R,int x,int z,int k,ll POW){
    int l = L, r = R;
    while (l <= r){
        int m = (l + r)>>1;
        ll tmp = POW + Pow(m,z) + (ll)x*m*z;
        if (tmp < 0) r = m  -1;//注意这里处理ll溢出时的操作wa好几次
        else if (tmp == k) return  true;
        else if (tmp < k) l = m + 1;
        else r = m - 1;
    }
    return false;
}
int main(){
   //freopen("din.txt","r",stdin);
    int K;
    while (~scanf("%d",&K)){
        if (!K) break;
        int ans = 0;
        for (int x = 1; x < 50000 && x < K; ++x){//枚举x
            for (int z = 2; z < 31; ++z){//枚举Z
                ll tmp = Pow(x,z);
                if (tmp > K) break;
                int mk = Bsearch(x + 1,50000,x,z,K,tmp);//二分查找是否存在满足的y值
                if (mk) ans++;
            }
        }
        printf("%d\n",ans);
    }
}