HDU1496Equations【hash】

Description

Consider equations having the following form: 

a*x1^2+b*x2^2+c*x3^2+d*x4^2=0 
a, b, c, d are integers from the interval [-50,50] and any of them cannot be 0. 

It is consider a solution a system ( x1,x2,x3,x4 ) that verifies the equation, xi is an integer from [-100,100] and xi != 0, any i ∈{1,2,3,4}. 

Determine how many solutions satisfy the given equation. 

 

Input

The input consists of several test cases. Each test case consists of a single line containing the 4 coefficients a, b, c, d, separated by one or more blanks. 
End of file.

 

Output

For each test case, output a single line containing the number of the solutions. 

 

Sample Input

1 2 3 -4 1 1 1 1

 

Sample Output

39088 0

 

分析：与上题相同

 

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int mod = 40007;

struct Node {
    int to;
    int next;
}e[mod + 10];

int head[mod + 10];

int tot;

void Add(int u, int v) {
    e[tot].to = v;
    e[tot].next = head[u];
    head[u] = tot++;
}

int Find(int u, int v) {
    int cnt = 0;
    for(int i = head[u]; i; i = e[i].next) {
        if(e[i].to == v) cnt++;
    }
    return cnt;
}

void init() {
    memset(head, 0, sizeof(head));
    tot = 1;
}

int Fabs(int x) {
    return x > 0 ? x : - x;
}

int pre[105];

int main() {
    int a, b, c, d;
    for(int i = 1; i <= 100; i++) {
        pre[i] = i * i;
    }
    while(EOF != scanf("%d %d %d %d",&a, &b, &c, &d) ) {
        init();
        for(int i = 1; i <= 100; i++) {
            for(int j = 1; j <= 100;    j++) {
                int num = a * pre[i] + b * pre[j];
                int p = Fabs(num) % mod;
                Add(p, num);
            }
        } 
        int ans = 0;
        for(int i = 1; i <= 100; i++) {
            for(int j = 1; j <= 100;    j++) {
                int num = c * pre[i] + d * pre[j];
                int p = Fabs(num) % mod;
                ans += Find(p, - num);
            }
        } 
        printf("%d
", ans << 4);
    }
    return 0;
}