sgu106.The equation 拓展欧几里得 难度:0

106. The equation

time limit per test: 0.25 sec. 
memory limit per test: 4096 KB

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

思路:
1 使用欧几里得构造出一组解使ax+by=gcd(a,b),然后(明显c%gcd!=0无解.)两边同乘以(c/gcd)
2 设k1=a/gcd,k2=b/gcd,(x,y)为原方程一组解,那么((x-n*k1),(y+n*k2))也是解(n为任意数)
3 于是不断寻找满足x1<=x<=x2,y1<=y<=y2的解,计数
4 这道题会爆int

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int inf=0x7ffffff;
long long tx1,tx2,ty1,ty2,a,b,c,tx,ty,minn,maxn;
void limit(long long L,long long R,long long d){//注意取区间端点
    if(d<0){L=-L;R=-R;d=-d;swap(R,L);}
    minn=max(minn,(long long)ceil((double)L/d));
    maxn=min(maxn,(long long)floor((double)R/d));
}
long long extgcd(long long a,long long b,long long &x,long long &y){
    long long d=a;
    if(b!=0){
        d=extgcd(b,a%b,y,x);
        y-=(a/b)*x;
    }
    else {
        x=1;y=0;
    }
    return d;
}
int main(){
    while(scanf("%I64d%I64d%I64d",&a,&b,&c)==3){
        scanf("%I64d%I64d%I64d%I64d",&tx1,&tx2,&ty1,&ty2);
        if(tx1>tx2||ty1>ty2){
            puts("0");continue;
        }
        long long ans=0;
        if(a==0&&b==0){
            if(c==0)ans=(tx2-tx1+1)*(ty2-ty1+1);
        }
        else if(a==0&&b){
            if(c%b==0&&(-c/b)>=ty1&&(-c/b)<=ty2){
              ans=(tx2-tx1+1);
            }
        }
        else if(b==0&&a){
            if(c%a==0&&(-c/a)>=tx1&&(-c/a)<=tx2){
            ans=(ty2-ty1+1);
            }
        }
        else {
            int d=extgcd(a,b,tx,ty);
            if((-c)%d==0){
                tx=-tx*c/d;
                ty=-ty*c/d;
                minn=-inf;maxn=inf;
                limit(tx1-tx,tx2-tx,b/d);
                limit(ty1-ty,ty2-ty,-a/d);
                if(minn<=maxn)ans=maxn-minn+1;
            }
        }
        printf("%I64d\n",ans);
    }
    return 0;
}