zoukankan      html  css  js  c++  java
  • POJ 1840 Eqs

    Eqs
    Time Limit: 5000MS   Memory Limit: 65536K
    Total Submissions: 15010   Accepted: 7366

    Description

    Consider equations having the following form: 
    a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0 
    The coefficients are given integers from the interval [-50,50]. 
    It is consider a solution a system (x1, x2, x3, x4, x5) that verifies the equation, xi∈[-50,50], xi != 0, any i∈{1,2,3,4,5}. 

    Determine how many solutions satisfy the given equation. 

    Input

    The only line of input contains the 5 coefficients a1, a2, a3, a4, a5, separated by blanks.

    Output

    The output will contain on the first line the number of the solutions for the given equation.

    Sample Input

    37 29 41 43 47

    Sample Output

    654

    Source


    按要求模拟即可

    hash是个神奇的东西

    下方代码注释部分是先三层循环后二层,正文部分是先二层循环后三层。两者都是正解,但是由于list插入比读取慢,先二层更快


     1 /*
     2 #include<iostream>
     3 #include<cstdio>
     4 #include<cstring>
     5 #include<algorithm>
     6 #include<list>
     7 using namespace std;
     8 const int mxn=14997;
     9 list<int>ha[mxn*2];
    10 list<int>::iterator it;
    11 int x,x1,x2,x3,x4,x5;
    12 int ans=0;
    13 int main(){
    14     scanf("%d%d%d%d%d",&x1,&x2,&x3,&x4,&x5);
    15     int i,j,k;
    16     for(i=-50;i<=50;i++)
    17       for(j=-50;j<=50;j++)
    18           for(k=-50;k<=50;k++){
    19               if(i==0||j==0||k==0)continue;
    20               x=i*i*i*x1+j*j*j*x2+k*k*k*x3;
    21               ha[x%mxn+mxn].push_back(x);//hash //x%mxn+mxn保证hash完以后是正数 
    22           }
    23     for(i=-50;i<=50;i++)
    24       for(j=-50;j<=50;j++){
    25           if(i==0|j==0)continue;
    26           x=-(i*i*i*x4+j*j*j*x5);
    27           //检查hash
    28         for(it=ha[x%mxn+mxn].begin();it!=ha[x%mxn+mxn].end();it++ ){
    29             if(*it==x)ans++;
    30         } 
    31       }
    32     printf("%d",ans);
    33     return 0;
    34     
    35 }
    36 */
    37 #include<iostream>
    38 #include<cstdio>
    39 #include<cstring>
    40 #include<algorithm>
    41 #include<list>
    42 using namespace std;
    43 const int mxn=14997;
    44 list<int>ha[mxn*2];
    45 list<int>::iterator it;
    46 int x,x1,x2,x3,x4,x5;
    47 int ans=0;
    48 int main(){
    49     scanf("%d%d%d%d%d",&x1,&x2,&x3,&x4,&x5);
    50     int i,j,k;
    51     for(i=-50;i<=50;i++)
    52       for(j=-50;j<=50;j++)
    53           {
    54               if(i==0||j==0)continue;
    55               x=i*i*i*x1+j*j*j*x2;
    56               ha[x%mxn+mxn].push_back(x);//hash //x%mxn+mxn保证hash完以后是正数 
    57           }
    58     for(i=-50;i<=50;i++)
    59       for(j=-50;j<=50;j++)
    60           for(k=-50;k<=50;k++){
    61               if(i==0|j==0||k==0)continue;
    62               x=-(i*i*i*x3+j*j*j*x4+k*k*k*x5);
    63               //检查hash
    64             for(it=ha[x%mxn+mxn].begin();it!=ha[x%mxn+mxn].end();it++ ){
    65                 if(*it==x)ans++;
    66             } 
    67           }
    68     printf("%d",ans);
    69     return 0;
    70     
    71 }




    niconiconi



    本文为博主原创文章,转载请注明出处。
  • 相关阅读:
    Servlet编程寄语
    filter常用功能
    Javascript的自动、定时执行和取消
    CentOS 5安装GIT的基本命令
    EF调用执行Oracle中序列
    WCF使用IIS发布服务的配置
    linux 自学系列:debian更新软件列表、更改源
    shell编程笔记五:select
    linux 自学系列: 改IP地址,主机名及DNS
    shell编程笔记四:case in
  • 原文地址:https://www.cnblogs.com/SilverNebula/p/5550601.html
Copyright © 2011-2022 走看看