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;
}