poj-1840-Eqs 哈希（hash）

来源：互联网发布：数组一般什么时候用编辑：程序博客网时间：2024/05/17 03:44

题目点我

Eqs

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 13400 Accepted: 6579

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

Source

Romania OI 2002

将公式变形，左边留下x1,x2, x3，这样可以先算左值，表示一个数组的下标，数组内存次数，

再计算右值，作为下标寻找次数，也就是映射，即哈希（hash）。

注意计算结果有正负，所以应将hash数组扩大一倍，并用short，否则会mle。

#include<stdio.h>#include<string.h>#define ll long long#define maxn 18750000short has[maxn*2+7];ll cnt=0;int main(){    //memset(has, sizeof(has), 0);    int a1, a2, a3, a4, a5;    short x1, x2, x3, x4, x5;    int ans;    scanf("%d%d%d%d%d", &a1, &a2, &a3, &a4, &a5);    for(x1=-50;x1<=50;x1++){        if(!x1) continue;        for(x2=-50;x2<=50;x2++){            if(!x2) continue;            for(x3=-50;x3<=50;x3++){                if(!x3) continue;                ans=a1*x1*x1*x1+a2*x2*x2*x2+a3*x3*x3*x3;                has[ans+maxn]++;            }        }    }    for(x4=-50;x4<=50;x4++){        if(!x4) continue;        for(x5=-50;x5<=50;x5++){            if(!x5) continue;            ans=-(a4*x4*x4*x4+a5*x5*x5*x5);            cnt+=has[ans+maxn];        }    }    printf("%I64u\n", cnt);    return 0;}

0 0