【POJ】1840：Eqs【哈希表】

Eqs

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 18299		Accepted: 8933

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

Source

Romania OI 2002

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Solution

一开始以为是meet in the middle搜索.....

然而完全没有那么复杂，甚至还可以用暴力map过？？

学习了一波hash表！

其实和建边的邻接表很像，就是把某些值系在某个特定的节点上，一般是定一个不大不小的模数来确定位置。

当然可能有重复，不过这就是hash表嘛！接在一起，查询就很接近$O(1)$了。

主要程序：

加入

void add(int v) {
	int x = v > 0 ? v : -v;
	x = (x % mod + x / mod) % mod;
	Edge[++stot] = Node(v, h[x]);
	h[x] = stot;
}

查询

int find(int v) {
	int ans = 0;
	int x = v > 0 ? v : -v;
	x = (x % mod + x / mod) % mod;
	for(int i = h[x]; i; i = Edge[i].nex)
		if(Edge[i].v == v)	ans ++;
	return ans;
}

很像邻接表吧~

mod是自己定的，这里定的100007，加入或查询都是按固定的mod方案就能固定位置了

#include<iostream>
#include<cstdio>
#define mod 1000007
using namespace std;

struct Node {
    int v, nex;
    Node() { }
    Node(int v, int nex) :
        v(v), nex(nex) { }
} Edge[1000010];

int stot, h[1000010];
void add(int v) {
    int x = v > 0 ? v : -v;
    x = (x % mod + x / mod) % mod;
    Edge[++stot] = Node(v, h[x]);
    h[x] = stot;
}

int find(int v) {
    int ans = 0;
    int x = v > 0 ? v : -v;
    x = (x % mod + x / mod) % mod;
    for(int i = h[x]; i; i = Edge[i].nex)
        if(Edge[i].v == v)    ans ++;
    return ans;
}

int main() {
    int a1, a2, a3, a4, a5;
    scanf("%d%d%d%d%d", &a1, &a2, &a3, &a4, &a5);
    for(int x1 = -50; x1 <= 50; x1 ++) if(x1)
        for(int x2 = -50; x2 <= 50; x2 ++) if(x2) {
            int s = x1 * x1 * x1 * a1 + x2 * x2 * x2 * a2;
            add(s);
        }
    int ans = 0;
    for(int x3 = -50; x3 <= 50; x3 ++) if(x3)
        for(int x4 = -50; x4 <= 50; x4 ++) if(x4)
            for(int x5 = -50; x5 <= 50; x5 ++) if(x5) {
                int s = x3 * x3 * x3 * a3 + x4 * x4 * x4 * a4 + x5 * x5 * x5 * a5;
                ans += find(s);
            }
    printf("%d", ans);
    return 0;
}