• The equation


    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

    大意:给你a,b,c,x1,y1,x2,y2。让你求ax+by+c=0在区间[x1,x2],[y1,y2]的整数解个数。
    考虑扩展欧几里得: 得到特解x0,y0后。设a=a/gcd,b=b/gcd。通解为x=x0+b*k,y=y0-a*k;
    则 得到不等式 x1<=x+b*k<=x2,y1<=y-a*k<=y2。
    解k的范围,取交集就好了。注意考虑向上取整与向下取整。
    比如下界就应该是想上取整,上界应该向下取整。
    #include <iostream>
    #include <algorithm>
    #define maxn 2025
    using namespace std;
    typedef long long ll;
    ll ans=0;
    ll L=-1e9,R=1e9;
    ll ex_gcd(ll a,ll b,ll &x,ll &y)
    {
        if(b==0)
        {
            x=1;
            y=0;
            return a;
        }
        ll ans=ex_gcd(b,a%b,x,y);
        ll temp=x;
        x=y;
        y=temp-(a/b)*x;
        return ans;
    }
    ll down(ll x,ll y)//向下取整
    {   ll res=0;
            res=x/y;
        if(x<0||y<0)
        {
            if(x%y!=0)
            {
                res--;
            }
        }
        return res;
    }
    ll upper(ll x,ll y)//向上取整
    {   ll res=x/y;
        if(x>0&&y>0||x<0&&y<0)
        {
            if(x%y!=0)
            {
                res++;
            }
        }
        return res;
    }
    void sovle(ll l,ll r,ll b)
    {
       if(b<0)
       {
           l=-l;
           r=-r;
           b=-b;
           swap(l,r);
       }
       L=max(L,upper(l,b));
       R=min(R,down(r,b));
    
    }
    int main()
    {
        int n;
        ll a,b,c,x1,y1,x2,y2;
        cin>>a>>b>>c>>x1>>x2>>y1>>y2;
        c=-c;
        if(a==0&&b==0&&c!=0)
        {
            ans=0;
        }
        else if(a==0&&b==0&&c==0)
        {
            ans=(x2-x1+1)*(y2-y1+1);
        }
        else if(a==0&&b!=0)
        {
            if(c%b==0)
            {
             ll  y=c/b;
                if(y>=y1&&y<=y2)
                {
                    ans++;
                }
            }
        }
        else if(b==0&&a!=0)
        {
            if(c%a==0)
            {
                ll x=c/a;
                if(x>=x1&&x<=x2)
                {
                    ans++;
                }
            }
        }
        else
        {   ll x,y;
            ll gcd=ex_gcd(a,b,x,y);
            if(c%gcd==0)
            {   x=x*c/gcd;
                y=y*c/gcd;
                a/=gcd;
                b/=gcd;
                ll lx=x1-x,rx=x2-x;
                ll ly=y1-y,ry=y2-y;
               sovle(lx,rx,b);
               sovle(ly,ry,-a);
               ans=R-L+1;
            }
        }
        cout<<ans<<endl;
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/zyf3855923/p/9016618.html
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