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  • Eqs (哈希)

    Time Limit: 5000MS   Memory Limit: 65536K
    Total Submissions: 10695   Accepted: 5185

    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
    题意:求出满足上述等式的解的个数,但x1,x2,x3,x4,x5不能为0;

    解题思路:第一感觉是五个循环枚举,那样果断会超时,其时我们可以把x1,x1移项,先枚举x1,x2,哈希之后再枚举x3,x4,x5;
    但-a1x1-a2x2结果有可能是负值,这时可以-a1x1-a2x2+12500000将结果变为正的;了解了传说中的哈希函数,值得学习;
     1 #include<stdio.h>
     2 #include<string.h>
     3 char hash[25000000];
     4 
     5 int main()
     6 {
     7     int a1,a2,a3,a4,a5;
     8     int x1,x2,x4,x5,x3,ans;
     9     memset(hash,0,sizeof(hash));
    10 
    11     scanf("%d %d %d %d %d",&a1,&a2,&a3,&a4,&a5);
    12 
    13     for(x1 = -50; x1 <= 50; x1++)
    14     {
    15         if(x1 == 0)continue;
    16         for(x2 = -50; x2 <= 50; x2++)
    17         {
    18             if(x2 == 0)continue;
    19             ans = -(x1*x1*x1*a1+x2*x2*x2*a2)+12500000;
    20             hash[ans]++;
    21         }
    22     }
    23     int cnt = 0;
    24     for(x3 = -50; x3 <= 50; x3++)
    25     {
    26         if(x3 == 0)continue;
    27         for(x4 = -50; x4 <= 50; x4++)
    28         {
    29             if(x4 == 0)continue;
    30             for(x5 = -50; x5 <= 50; x5++)
    31             {
    32                 if(x5 == 0)continue;
    33                 ans = x3*x3*x3*a3+x4*x4*x4*a4+x5*x5*x5*a5+12500000;
    34                 if(ans >= 0 && ans < 25000000)
    35                     cnt += hash[ans];
    36             }
    37         }
    38     }
    39     printf("%d
    ",cnt);
    40     return 0;
    41 }
    View Code
    
    
     
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  • 原文地址:https://www.cnblogs.com/LK1994/p/3257396.html
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