（light oj 1306） Solutions to an Equation 扩展欧几里得算法

题目链接：http://lightoj.com/volume_showproblem.php?problem=1306

You have to find the number of solutions of the following equation:

Ax + By + C = 

Where A, B, C, x, y are integers and x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2.

Input
Input starts with an integer T (≤ ), denoting the number of test cases.

Each case starts with a line containing seven integers A, B, C, x1, x2, y1, y2 (x1 ≤ x2, y1 ≤ y2). The value of each integer will lie in the range [-, ].

Output
For each case, print the case number and the total number of solutions.
Sample Input

  - -
- -  -  -
  -
- -  -  -
  -
Output for Sample Input
Case :
Case :
Case :
Case :
Case : 

题意：给出AX+BY+C==0中的A，B，C。问在X1到X2与Y1到Y2的范围内有几组解

分析：利用扩展欧几里得算法

首先我们可以求出ax+by=gcd(a,b)=g的一个组解(x₀,y₀).而要使ax+by=c有解，必须有c%g==0.

继而可以得到ax+by=c的一个组解x1=c*x₀/g , y1=c*y₀/g。

这样可以得到ax+by=c的通解为：

                  x=x1+b*t;

                  y=y1-a*t;

再就是要注意符号问题！！！

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stack>
#include <map>
#include <string>
#include <vector>
#include<iostream>
using namespace std;
#define N 10006
#define INF 0x3f3f3f3f
#define LL long long
#define mod 1000000007
LL ex_gcd(LL a,LL b,LL &x,LL &y)
{
    if(b==)
    {
        x = ;
        y = ;
        return a;
    }
    LL g = ex_gcd(b,a%b,x,y);
    LL t = x;
    x = y;
    y = t- a/b * y;
    return g;
}
int sign(LL n)
{
    if(n==)
        return ;
    return n>?:-;
}
LL ceil(LL a,LL b)
{
    int s = sign(a) * sign(b);
    return b/a + (b%a!= && s>);
}
LL floor(LL a,LL b)
{
    int s = sign(a) * sign(b);
    return b/a - (b%a!= && s<);
}
int main()
{
    int T,con=;
    scanf("%d",&T);
    LL a,b,c,x1,x2,y1,y2,x,y;
    while(T--)
    {
        scanf("%lld %lld %lld %lld %lld %lld %lld",&a,&b,&c,&x1,&x2,&y1,&y2);
        printf("Case %d: ",con++);
        if(a== && b==)
        {
            if(c==)
            {
                printf("%lld\n",(x2-x1+)*(y2-y1+));
            }
            else
                printf("0\n");
            continue;
        }
        if(a==)
        {
            if(c%b!=)
            {
                printf("0\n");
                continue;
            }
            LL s = -c/b;
            if(s>=y1 && s<=y2)
                printf("%lld\n",x2-x1+);
            else
                printf("0\n");
            continue;
        }
        if(b==)
        {
            if(c%a!=)
            {
                printf("0\n");
                continue;
            }
            LL s = -c/a;
            if(s>=x1 && s<=x2)
                printf("%lld\n",y2-y1+);
            else
                printf("0\n");
            continue;
        }

        LL g = ex_gcd(a,b,x,y);
        if(c%g!=)
        {
            printf("0\n");
            continue;
        }
        if(sign(g) * sign(b) <) swap(x1,x2);
        LL l1 = ceil(b, g*x1 + c*x);
        LL l2 = floor(b, g*x2 + c*x);
        if(sign(-a) * sign(g) <) swap(y1,y2);
        LL r1 = ceil(-a,g * y1 + c*y);
        LL r2 = floor(-a,g*y2 + c*y);
        l1 = max(l1,r1);
        r1 = min(l2,r2);
        if(l1>r1) printf("0\n");
        else
            printf("%lld\n",r1-l1 +);
    }
    return ;
}