hdu 4282 A very hard mathematic problem(二分)

EEGLAB安装、EDF数据导入、EEG预处理(Matlab) 目录下载和安装下载安装遇到问题及解决方案导入EDF数据EEG预处理 下载和安装 下载 链接:https://pan.baidu.com/s/17eMT6B7LgZ4PyqWvD5ixmg 提取码:12gz 安装 **第一步:**将下载下来的文件解压,修改名称为eeglab,将此文件夹存放在matlab的安装目录里的toolbox文件夹下,下图是我的路径,需要拷贝到自己相对应的路径中 第二步: 打开Matlab软件,点击主页中的设置路径 设置路径 点击添加并包含子文件夹,将eeglab文件夹添加进来,保 阅读详情
  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
9
53
6
0
 
Sample Output
1
1
0   
Hint
9 = 1^2 + 2^2 + 1 * 2 * 2
53 = 2^3 + 3^3 + 2 * 3 * 3
http://acm.hdu.edu.cn/showproblem.php?pid=4282
题意:给定 K ,求满足 X^Z + Y^Z + XYZ = K 这个等式的解。

分析:f(x)=x^b + a*b*x + a^b - k;枚举a 和b,二分找  x (x的范围[a+1,pow(k,1/b)]);

#include<iostream>
#include<cstdio>
#include<cmath>

using namespace std;

int k,maxa;

inline int f(int a,int b,int x)
{
    return pow(x*1.0,b*1.0)+pow(a*1.0,b*1.0)+a*b*x-k;
}

bool judge(int a,int b)
{
    int L,R,M,t;
    L=a+1;R=maxa;
    if(f(a,b,L)>0||f(a,b,R)<0) return false;
    while(L<=R)
    {
        M=(L+R)>>1;
        t=f(a,b,M);
        if(t==0) return true;
        else if(t>0) R=M-1;
        else L=M+1;
    }
    return false;
}

int main()
{
    int ans,a,b;
    while(scanf("%d",&k)==1&&k)
    {
        ans=0;
        for(b=2;b<=30;b++)
        {
            maxa=pow(k*1.0,1.0/b);
            for(a=1;a<=maxa;a++)
            {
                if(judge(a,b))
                    ans++;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}
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HDU 4282 A very hard mathematic problem (暴力枚举+二分+各种剪枝)

A very hard mathematic problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 8579    Accepted Submission(s): 2624   Problem Description   ...

LSD20164388的博客 593

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...

weixin_30284355的博客 101

hdu4282 A very hard mathematic problem(二分)

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

qq_45928501的博客 139

A very hard mathematic problem二分HDU4282

#include<iostream> #include<algorithm> #include<cstdio> #include<string> #include<vector> #include<string.h> #include<map> #include<cmath> #include<queue> #define ll long long #define INF 0x7fffffff #de

the chosen 07 的博客 130
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