LightOJ 1306 Solutions to an Equation

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

5

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 longusing 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）