codevs1213 解的个数

题目描述 Description

已知整数x,y满足如下面的条件:

 

ax+by+c = 0

p<=x<=q

r<=y<=s

 

求满足这些条件的x,y的个数。

输入描述 Input Description

第一行有一个整数n（n<=10），表示有n个任务。n<=10

以下有n行，每行有7个整数，分别为：a,b,c,p,q,r,s。均不超过108。

输出描述 Output Description

共n行，第i行是第i个任务的解的个数。

样例输入 Sample Input

2

2 3 -7 0 10 0 10

1 1 1 -10 10 -9 9

样例输出 Sample Output

1

19

#include<iostream>
#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;
ll x,y,a,b,c,g,p,q,r,s;
ll exgcd(ll a,ll b,ll &x,ll &y){
    if(b == 0){
        x = 1;
        y = 0;
        return a;
    }
    ll ans = exgcd(b,a%b,x,y);
    ll t = x;
    x = y;
    y = t - (a/b) * y;
    return ans;
}
ll gcd(ll a,ll b){
    return b == 0 ? a : gcd(b,a%b);
}
int main(){
    int n;
    cin>>n;
    for(int i = 1;i <= n;i++){
    cin>>a>>b>>c>>p>>q>>r>>s;
    c = -c;
    if(a == 0 && b == 0 && c!= 0){
        cout<<0<<endl;
        continue;
    }
    if(q < p || s < r){
        cout<<0<<endl;
        continue;
    }
    if(a == 0 && b== 0 && c==0){
        cout<<(q-p+1)*(s-r+1)<<endl;
        continue;
    }
    g = exgcd(a,b,x,y);
    x *= c/g;
    y *= c/g;
    if(x*a + y*b != c){
        cout<<0<<endl;
        continue;
    }
    ll plusx = b / g,plusy = a / g,dx,dy;
    ll acc = 0;
    bool fu;
    if(plusx < 0) fu = true;
    else fu = false;
    while(x < p){
        if(plusx == 0) break;
        if(!fu) x+=plusx,y-=plusy;
        else x-=plusx,y+=plusy;
    }
    while(x > q){
        if(plusx == 0) break;
        if(!fu) x-= plusx,y+=plusy;
        else x+=plusx,y-=plusy;
    }
    if(x >= p && x <= q && y >= r && y <= s)acc++;
    dx = x;
    dy = y;
    while(dx >= p){
        if(plusx == 0) break;
        if(!fu){
            dx -= plusx;
            dy += plusy;
        }else{
            dx += plusx;
            dy -= plusy;
        }
        if(dx >= p && dx <= q && dy >= r && dy <= s) acc++;
    }
    dx = x;
    dy = y;
    while(dx <= q){
        if(plusx == 0) break;
        if(!fu){
            dx += plusx;
            dy -= plusy;
        }else{
            dx -= plusx;
            dy += plusy;
        }
        if(dx >= p && dx <= q && dy >= r && dy <= s) acc++;
    }
    cout<<acc<<endl;
    }
    return 0;
}