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  • poj 1840 暴力+标记

    Description

    Consider equations having the following form: 
    a1x1 3+ a2x2 3+ a3x3 3+ a4x4 3+ a5x5 3=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

    暴力+标记
     1 #include <cstdio>
     2 #include <cmath>
     3 #include <cstring>
     4 #include <ctime>
     5 #include <iostream>
     6 #include <algorithm>
     7 #include <set>
     8 #include <vector>
     9 #include <sstream>
    10 #include <queue>
    11 #include <typeinfo>
    12 #include <map>
    13 #include <stack>
    14 #define inf 0xfffff
    15 typedef long long ll;
    16 using namespace std;
    17 #define MAXN 25000000
    18 #define mod 10007
    19 #define eps 1e-9
    20 short hash[25000001];
    21 int main()
    22 {
    23     int a1,a2,a3,a4,a5;
    24     while(scanf("%d %d %d %d %d",&a1,&a2,&a3,&a4,&a5)!=EOF)
    25     {
    26         int sum=0;
    27         for(int i=-50; i<=50; i++)
    28             for(int j=-50; j<=50; j++)
    29                 for(int k=-50; k<=50; k++)
    30                 {
    31                     if(i==0||j==0||k==0)
    32                         continue;
    33                     if(a1*i*i*i+a2*j*j*j+a3*k*k*k<0)
    34                         hash[a1*i*i*i+a2*j*j*j+a3*k*k*k+MAXN]++;
    35                     else
    36                         hash[a1*i*i*i+a2*j*j*j+a3*k*k*k]++;
    37                 }
    38 
    39         for(int i=-50; i<=50; i++)
    40             for(int j=-50; j<=50; j++)
    41             {
    42                 int u=0-a4*i*i*i-a5*j*j*j;
    43                 if(u<0)
    44                     u=u+MAXN;;
    45                 if(hash[u]&&i!=0&&j!=0)
    46                 {
    47                     sum+=hash[u];
    48                 }
    49             }
    50         printf("%d
    ",sum);
    51     }
    52     return 0;
    53 }
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  • 原文地址:https://www.cnblogs.com/hsd-/p/4655942.html
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