zoukankan      html  css  js  c++  java
  • poj 1840 -- Eqs

    Eqs
    Time Limit: 5000MS   Memory Limit: 65536K
    Total Submissions: 11865   Accepted: 5811

    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

    思路:hash水过。

     1 /*======================================================================
     2  *           Author :   kevin
     3  *         Filename :   Eqs.cpp
     4  *       Creat time :   2014-07-24 17:49
     5  *      Description :
     6 ========================================================================*/
     7 #include <iostream>
     8 #include <algorithm>
     9 #include <cstdio>
    10 #include <cstring>
    11 #include <queue>
    12 #include <cmath>
    13 #define clr(a,b) memset(a,b,sizeof(a))
    14 #define M 12500000
    15 using namespace std;
    16 short int ha[2*M+5];
    17 int main(int argc,char *argv[])
    18 {
    19     int a,b,c,d,e;
    20     while(scanf("%d%d%d%d%d",&a,&b,&c,&d,&e)!=EOF){
    21         clr(ha,0);
    22         for(int i = -50; i <= 50; i++){
    23             if(!i) continue;
    24             for(int j = -50; j <= 50; j++){
    25                 if(!j) continue;
    26                 int s = a*i*i*i+b*j*j*j;
    27                 ha[M-s]++;
    28             }
    29         }
    30         int cnt = 0;
    31         for(int i = -50; i <= 50; i++){
    32             if(!i) continue;
    33             for(int j = -50; j <= 50; j++){
    34                 if(!j) continue;
    35                 for(int k = -50; k <= 50; k++){
    36                     if(!k) continue;
    37                     int s = c*i*i*i+d*j*j*j+e*k*k*k;
    38                     s = M+s;
    39                     if(s < 2*M && s > 0)
    40                         cnt += ha[s];
    41                 }
    42             }
    43         }
    44         printf("%d
    ",cnt);
    45     }
    46     return 0;
    47 }
    View Code
  • 相关阅读:
    pku3225 区间
    pku2136 Vertical Histogram
    NOI2006 最大获利
    APIO2010 特别行动队
    停电两夜
    偷偷乐一把
    那些花儿...
    一件一块钱的小事
    web service 的Section=ResponseStatusLine 错误和skype
    杂谈
  • 原文地址:https://www.cnblogs.com/ubuntu-kevin/p/3866389.html
Copyright © 2011-2022 走看看