HDU_4282

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A very hard mathematic problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6957 Accepted Submission(s): 2079

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.

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.
　　

Output

　　Output the total number of solutions in a line for each test case.

Sample Input

Sample Output

#include <iostream>#include <cstdio>#include <cstring>#define ll long longusing namespace std;ll num[50001][33] = {0};void Init(){    for(ll i = 1; i < 50001; i++)    {        num[i][1] = i;        for(int j = 2; j < 33; j++)        {            num[i][j] = num[i][j - 1] * i;            if(num[i][j] > 2147483648LL)            {                break;            }        }    }}bool judge(ll x, ll z, ll snum){    ll r = x + 1, l = 50000;    for(;r <= l;)//二分判断找Y    {        ll mid = (l + r)>>1;        if(num[mid][z] == 0)//这个地方一直没有加上，一直WA，没意识到y^z越界的问题        {            l = mid - 1;            continue;        }        if(num[mid][z] + x * z * mid < snum)        {            r = mid + 1;        }        else if(num[mid][z] + x * z * mid > snum)        {            l = mid - 1;        }        else        {            //printf("%d^%d + %d^%d + %d*%d*%d\n", x, z, mid, z, x, mid, z);            return true;        }    }    return false;}int main(){    ll k;    Init();    for(;scanf("%I64d", &k),k;)//以输入的数为判断条件的循环输入方法    {        ll res = 0;        for(ll x = 1; x <= k && x < 50001; x++)        {            for(ll z = 2; z < 32; z++)//一开始没注意题目要求x < y && z > 1 =_=            {                if(num[x][z] == 0)//超过2^32部分舍弃                {                    break;                }                ll snum = k - num[x][z];                if(snum - x * z * x< 0)//不存在这样的Y                {                    break;                }                if(judge(x, z, snum))                {                    res++;                }            }        }        printf("%d\n", res);    }    return 0;}</cstring></cstdio></iostream>

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