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

View Code

#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);

    }

}