LightOJ 1306 Solutions to an Equation

最新推荐文章于 2019-02-01 12:54:57 发布

原创最新推荐文章于 2019-02-01 12:54:57 发布 · 612 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#扩欧 #Solutions to an Equa

扩展欧几里得专栏收录该内容

6 篇文章

订阅专栏

LightOJ 1306 Solutions to an Equation

You have to find the number of solutions of the following equation:

Ax + By + C = 0

Where A, B, C, x, y are integers and x₁ ≤ x ≤ x₂ andy₁ ≤ y ≤ y₂.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case starts with a line containing seven integers A, B, C, x₁, x₂, y₁, y₂ (x₁ ≤ x₂, y₁ ≤ y₂). The value of each integer will lie in the range [-10⁸, 10⁸].

Output

For each case, print the case number and the total number of solutions.

Sample Input

1 1 -5 -5 10 2 4

-10 -8 80 -100 100 -90 90

2 3 -4 1 7 0 8

-2 -3 6 -2 5 -10 5

1 8 -32 0 0 1 10

Sample Output

Case 1: 3

Case 2: 37

Case 3: 1

Case 4: 2

Case 5: 1

题意：给你一个方程求有多少组解。给定的数正负零都可以。

巨怕这种题（orz）

思路：扩欧的应用，注意正负零的特判，还有其他的一些情况

#pragma comment(linker, "/STACK:1024000000,1024000000")
//#include <bits/stdc++.h>
#include<string>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
#define maxn 10010
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MOD 1000000007
#define ll long long
using namespace std;

ll exgcd(ll a, ll b, ll &x, ll &y)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    ll d = exgcd(b, a % b, y, x);
    y -= (a / b)*x;
    return d;
}
 
int main()
{
    int T;
    int cnt = 0;
    cin >> T;
    while(T--)
    {
        ll a, b, c, x1, x2, y1, y2;
        scanf("%lld%lld%lld", &a, &b, &c);
        scanf("%lld%lld%lld%lld",&x1, &x2, &y1, &y2);
        c = -c;
        if(a < 0)
        {
            a = -a;
            swap(x1, x2);
            x1 = -x1;
            x2 = -x2;
        }
        if(b < 0)
        {
            b = -b;
            swap(y1, y2);
            y1 = -y1;
            y2 = -y2;
        }
        printf("Case %d: ", ++cnt);
       
        ll x0 = 0 , y0 = 0;
        ll g = exgcd(a, b, x0, y0);
        if(g!=0 && c % g != 0)
        {
            printf("0\n");
            continue;
        }
        if(a == 0 && b == 0)
        {
            if(!c)
                printf("%lld\n", (x2 - x1 + 1) * (y2 - y1 + 1));
            else 
                printf("0\n");
            continue;
        }
        if(a == 0)
        {
            if(c / b >= y1 && c / b <= y2)
                printf("%lld\n", x2 - x1 + 1);
            else 
                printf("0\n");
            continue;
        }
        if(b == 0)
        {
            if(c / a >= x1 && c / a <= x2)
                printf("%lld\n", y2 - y1 + 1);
            else 
                printf("0\n");
            continue;
 
        }
        x0 = x0 * c / g;
        y0 = y0 * c / g;
        a /= g;
        b /= g;
        ll l = ceil((double)(x1 - x0)/(double)(b));
        ll r = floor((double)(x2 - x0)/(double)(b));
        ll w = ceil((double)(y0 - y2)/(double)(a));
        ll e = floor((double)(y0 - y1)/(double)(a));
        ll q = max(l, w);
        ll p = min(r, e);
        if(p < q)
            printf("0\n");
        else printf("%lld\n", p - q + 1);
    }
    return 0;
}

//向上取整（ceil()） 向下取整（floor）