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  • 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
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  • 原文地址:https://www.cnblogs.com/ubuntu-kevin/p/3866389.html
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