sgu The equation

There is an equation ax + by + c = 0. Given a,b,c,x1,x2,y1,y2 you must determine, how many integer roots of this equation are satisfy to the following conditions : x1<=x<=x2,   y1<=y<=y2. Integer root of this equation is a pair of integer numbers (x,y).

Input

Input contains integer numbers a,b,c,x1,x2,y1,y2 delimited by spaces and line breaks. All numbers are not greater

算法：

    ax + by + c = 0

   因为，扩展欧几里得处理不了数为负数的情况。所以，要保障a,b,c为整数。下面分析过程：

1、(-a)x + (-b)y = c  需要判断a,b,c的正负值。

     c < 0 -----> a = -a,b = -b 取值转变区间[x1,x2]------> [x2,x1] ,[y1,y2] -------->[y2,y1]

    a < 0 ------>直接变区间

    b < 0 同理

2、判断ax + by = -c ，其中a,b是否为0的情况。因为，这时候是比较特殊的。

   1.a = 0 && b = 0 --->c = 0 有解，否则无解

   2.a = 0&&b != 0 --> by = -c --->y = -c/b判断  -c % b == 0 && y1 <= y <= y2 ？成立有解，否则无解

   3.b = 0&&b ！=0 同理

3,ax + by = -c ，其中a != 0 && b != 0，运用扩展欧几里得。求出其中|x| + |y| 的最小解，然后再判断。

    x = x0 + kb --> k = (x - x0) / b ----> x1 <= k <= x2 

    y = y0 - kb ---> k=(y-y0)/b ----> y2 <= k <= y1 （注意负号使得区间转换！！！！！）

   最后，取区间的交集就可以了。

#include <iostream>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

typedef long long LL;
typedef pair<int,int> P;

inline int read(){
    int x = 0,f = 1; char ch = getchar();
    while(ch < '0'||ch > '9'){if(ch == '-')f=-1;ch = getchar();}
    while(ch >= '0'&&ch <= '9'){x = x * 10 + ch -'0';ch = getchar();}
    return x*f;
}

//

inline long long cmin(const long long &x,const long long &y) {return x<y?x:y;}
inline long long cmax(const long long &x,const long long &y) {return x>y?x:y;}

LL gcd(LL a,LL b){
    return b ? gcd(b,a%b) : a;
}

void extgcd(LL a,LL b,LL& d,LL& x,LL& y){
    if(!b) { d = a; x = 1;y = 0;}
    else { extgcd(b,a%b,d,y,x); y -= x*(a/b);}
}

int main()
{
    LL a,b,c,x1,x2,y1,y2;
    cin >> a >> b >> c >> x1 >> x2 >> y1 >> y2;

    c = -c;
    if(c < 0) {a = -a; b = -b; 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;}

    if(a == 0&&b == 0){
        if(c == 0){
            printf("%I64d",(x2-x1+1)*(y2-y1+1));
        } else {
            printf("0");
        }
       return 0;
    }

    if(a == 0){
        if(c % b == 0&&y1 <= c/b&&c/b <= y2){
            printf("%I64d",x2-x1+1);
        } else {
           printf("0");
        }
        return 0;
    }

    if(b == 0){
        if(a % c == 0 && x1 <= c/a&&c/a <= x2){
            printf("%I64d",y2-y1+1);
        } else {
            printf("0");
        }
       return 0;
    }

    LL d,x,y;
    d = gcd(a,b);

    if(c%d){
        printf("0");
       return 0;
    }

    a /= d; b /= d; c /= d;

    extgcd(a,b,d,x,y);

    x *= d; y *= d;

    double tx1 = x1,tx2 = x2,ty1 = y1,ty2 = y2,tb = b,ta = a,tx = x,ty = y;
    LL downx = floor((tx2 - tx) / tb),downy = floor((ty2 - ty) / ta);
    LL upx = ceil((tx1 - tx) / tb),upy = floor((ty1 - ty) / ta);
    LL r = cmin(downx,downy);
    LL l = cmax(upx,upy);
    LL ans;
    if(r < l) ans = 0;
    else ans = r - l + 1;

    printf("%I64d",ans);
    return 0;
}



/*

1 1 -3
0 4
0 4

*/

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