SGU106 The equation[扩展欧几里德算法]

A - The equation

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

Submit  Status

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

题意:

求出 ax+by=-c中的解 x y  符合    x1<=x<=x2  y1<=y<=y2 的解的个数

自己真的挺难理解这个的..然后参考了两个博客

都写的很详细. 直接贴出来去看吧..我感觉自己没这个水平去讲这个

http://blog.youkuaiyun.com/volzkzg/article/details/7427233

http://www.cnblogs.com/Rinyo/archive/2012/11/25/2787419.html

主要用到了扩展欧几里德算法  变形一下, 这些博客里面都有说道

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<vector>
using namespace std;
long long gcd(long long a,long long b)
{
    return b==0?a:gcd(b,a%b);
}
long long exgcd(long long a,long long b,long long &x,long long &y)
{
    if (!b)
    {
        x=1;
        y=0;
        return a;
    }
    long long gcd=exgcd(b,a%b,x,y);
    long long temp=x;
    x=y;
    y=temp-a/b*y;
    return gcd;
}
long long upper(long long a,long long b)//往上取整
{
    if (a<=0)
        return a/b;
    return (a-1)/b+1;
}
long long lower(long long a,long long b)//往下取整
{
    if (a>=0)
        return a/b;
    return (a+1)/b-1;
}
long long get(long long l,long long r,long long d,long long &k1,long long &k2)
{
    if (d<0)
    {
        l=-l;
        r=-r;
        d=-d;
        swap(l,r);
    }
    k1=upper(l,d);
    k2=lower(r,d);
}
int main()
{
    long long a,b,c,x1,y1,x2,y2;
    while (cin>>a>>b>>c)//ax+by+c=0 >>ax+by=-c
    {
        bool flag=0;
        c=-c;
        cin>>x1>>x2>>y1>>y2;
        long long ans,x,y;
        if (!a && !b && !c)
        {
            cout<<(x2-x1+1)*(y2-y1+1)<<endl;
            continue;
        }
        else if (!a && !b)
        {
            cout<<0<<endl;
            continue;
        }
        else if (!a)
        {
            if (c%b || c/b<y1 || c/b>y2)
                ans=0;
            else
                ans=x2-x1+1;
            cout<<ans<<endl;
            continue;
        }
        else if (!b)
        {
            if (c%a || c/a<x1 || c/a>x2)
                ans=0;
            else
                ans=y2-y1+1;
            cout<<ans<<endl;
            continue;
        }
        /*
        先处理了a,b,c为0的情况
        */

        long long g=gcd(a,b);
        if (c%g)//如果 c不是gcd(a,b)的倍数 无解
        {
            cout<<0<<endl;
            continue;
        }
        /*
        网上的解释:
        方程两边同时除以gcd(a,b).我们假设aa=a/gcd(a,b),bb=b/gcd(a,b),nn=n/gcd(a,b)
        所以方程两边同时除以gcd(a,b)后，
        可以得到一个方程aa*x+bb*y=nn.
        并且该方程aa*x+bb*y=nn的解x,y就是a*x+b*y=n的解
        我们只要求解出aa*x+bb*y=1的其中一个解，设这两个解为x0,y0.
        那么aa*x+bb*y=nn的其中一个解解就是x0*nn,y0*nn.
        接着，a*x+b*y=n的其中一个解解也就是x0*nn,y0*nn.
        a*(x0*nn)+b*(y0*nn)=n.
        我们会发现
        a*(x0*nn+1*b)+b*(y0*nn-1*a)=n
        a*(x0*nn-1*b)+b*(y0*nn+1*a)=n.
        继续推广
        a*(x0*nn+k*b)+b*(y0*nn-k*a)=n （k属于整数）
        nn=n/gcd(a,b).
        x=x0*nn+k*b
        y=y0*nn-k*a
        */
        a/=g;
        b/=g;
        c/=g;// c=c/g 上面已经将c变为-c
        exgcd(a,b,x,y);
        long long k1,k2,k3,k4;
        x*=c;
        y*=c;
        get(x1-x,x2-x,b,k1,k2);
        get(y1-y,y2-y,-a,k3,k4);
//        cout<<x<<" "<<y<<endl;
        ans=min(k2,k4)-max(k1,k3)+1;
        cout<<ans<<endl;
    }
    return 0;
}