SGU 106 The equation

H - The equation

Time Limit:250MS     Memory Limit:4096KB     64bit IO Format:%I64d & %I64u

Submit Status Practice SGU 106

Description

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 than 10⁸ by absolute value。

Output

Write answer to the output.

Sample Input

1 1 -3
0 4
0 4

Sample Output

4

我的思路就是首先把一个基本解求出来，然后看在x1、x2的范围内x的范围是多少，然后找到对应的y的范围，再看y的范围有多少个解是在y1、y2范围之内的，这个就是最后的答案。

    当然，对于含有a=0或b=0的情况要特判一下。

附上一个很不错的网址：传送门

#include <iostream>
using namespace std;
typedef long long LL;
LL a,b,c,x1,x2,y1,y2,x,y,tmp,ans=0;
LL mini = -361168601842738790LL;
LL maxi = 322337203685477580LL;
int extendedGcd(int a,int b){
    if (b==0){
    x=1;y=0;
    return a;
    }
    else{
    int tmp = extendedGcd(b,a%b);
    int t = x;
    x=y;
    y=t-a/b*y;
    return tmp;
    }
}
LL extendedGcd(LL a,LL b){
    if (b == 0){
    x=1;y=0;
    return a;
    }
    else{
    LL TEMP = extendedGcd(b,a%b);
    LL tt=x;
    x=y;
    y=tt-a/b*y;
    return TEMP;
    }
}
LL upper(LL a,LL b){
    if (a<=0)
    return a/b;
    return (a-1)/b + 1;
}
LL lower(LL a,LL b){
    if (a>=0)
    return a/b;
    return (a+1)/b - 1;
}
void update(LL L,LL R,LL wa){
    if (wa<0){
    L=-L;R=-R;wa=-wa;
    swap(L,R);
    }
    mini=max(mini,upper(L,wa));
    maxi=min(maxi,lower(R,wa));
}
int main(){
    cin >> a >> b >> c >> x1 >> x2 >> y1 >> y2;c=-c;
    if (a==0 && b==0){
    if (c==0) ans = (x2-x1+1) * (y2-y1+1);
    }
    else if (a==0 && b!=0){
    if (c % b==0) {
        tmp = c/b;
        if (tmp>=y1 && tmp<=y2) ans = 1;
    }
    }
    else if (a!=0 && b==0){
    if (c % a==0){
        tmp = c/a;
        if (tmp>=x1 && tmp<=x2) ans = 1;
    }
    }
    else{
    LL d = extendedGcd(a,b);
    if (c%d == 0){
        LL p = c/d;
        update(x1-p*x,x2-p*x,b/d);
        update(y1-p*y,y2-p*y,-a/d);
        ans = maxi-mini+1;
        if (ans<0) ans=0;
    }
    }
    cout << ans << endl;
}