zoukankan      html  css  js  c++  java
  • SGU 106 The equation 扩展欧几里德

    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:ax+by=-c;
       扩展欧几里德求解;
       x=x0+b/gcd(a,b)*t;
       y=y0+a/gcd(a,b)*t;
    求x1<=x<=x2&&y1<=y<=y2的条件下,t的可行解;
       找到x的范围的t的可行解[lx,rx];
       同理 [ly,ry];
    ans=min(rx,ry)-max(lx,ly)+1;
    #include<bits/stdc++.h>
    using namespace std;
    #define ll __int64
    #define esp 1e-13
    const int N=1e3+10,M=1e6+1000,inf=1e9+10,mod=1000000007;
    void extend_Euclid(ll a, ll b, ll &x, ll &y)
    {
        if(b == 0)
        {
            x = 1;
            y = 0;
            return;
        }
        extend_Euclid(b, a % b, x, y);
        ll tmp = x;
        x = y;
        y = tmp - (a / b) * y;
    }
    ll gcd(ll a,ll b)
    {
        if(b==0)
            return a;
        return gcd(b,a%b);
    }
    int main()
    {
        ll a,b,c;
        ll lx,rx;
        ll ly,ry;
        scanf("%I64d%I64d%I64d",&a,&b,&c);
        scanf("%I64d%I64d",&lx,&rx);
        scanf("%I64d%I64d",&ly,&ry);
        c=-c;
        if(lx>rx||ly>ry)
        {
            printf("0
    ");
            return 0;
        }
        if (a == 0 && b == 0 && c == 0)
        {
            printf("%I64d
    ",(rx-lx+1) * (ry-ly+1));
            return 0;
        }
        if (a == 0 && b == 0)
        {
            printf("0
    ");
            return 0;
        }
        if (a == 0)
        {
            if (c % b != 0)
            {
                printf("0
    ");
                return 0;
            }
            ll y = c / b;
            if (y >= ly && y <= ry)
            {
                printf("%I64d
    ",rx - lx + 1);
                return 0;
            }
            else
            {
                printf("0
    ");
                return 0;
            }
        }
        if (b == 0)
        {
            if (c % a != 0)
            {
                printf("0
    ");
                return 0;
            }
            ll x = c / a;
            if (x >= lx && x <= rx)
            {
                printf("%I64d
    ",ry - ly + 1);
                return 0;
            }
            else
            {
                printf("0
    ");
                return 0;
            }
        }
        ll hh=gcd(abs(a),abs(b));
        if(c%hh!=0)
        {
            printf("0
    ");
            return 0;
        }
        else
        {
            ll x,y;
            extend_Euclid(abs(a),abs(b),x,y);
            x*=(c/hh);
            y*=(c/hh);
            if(a<0)
                x=-x;
            if(b<0)
                y=-y;
            a/=hh;
            b/=hh;
            ll tlx,trx,tly,trry;
            if(b>0)
            {
                ll l=lx-x;
                tlx=l/b;
                if(l>=0&&l%b)
                    tlx++;
                ll r=rx-x;
                trx=r/b;
                if(r<0&&r%b)
                    trx--;
            }
            else
            {
                b=-b;
                ll l=x-rx;
                tlx=l/b;
                if(l>=0&&l%b)
                    tlx++;
                ll r=x-lx;
                trx=r/b;
                if(r<0&&r%b)
                    trx--;
            }
            if(a>0)
            {
                ll l=-ry+y;
                tly=l/a;
                if(l>=0&&l%a)
                    tly++;
                ll r=-ly+y;
                trry=r/a;
                if(r<0&&r%a)
                    trry--;
            }
            else
            {
                a=-a;
                ll l=ly-y;
                tly=l/a;
                if(l>=0&&l%a)
                    tly++;
                ll r=ry-y;
                trry=r/a;
                if(r<0&&r%a)
                    trry--;
            }
            printf("%I64d
    ",(max(0LL,min(trry,trx)-max(tly,tlx)+1)));
            return 0;
        }
        return 0;
    }

  • 相关阅读:
    Building a ListBox with custom content in Silverlight 4.0
    asp.net通讯问题
    Using the NavigationService Object in SL4.0
    Creating a File Explorer for Isolated Storage
    图表ASP:Chart
    什么是继承?
    Java基础一笔带过
    Java多态
    自己动手写个小框架之七
    linux crontab 定时计划
  • 原文地址:https://www.cnblogs.com/jhz033/p/5755075.html
Copyright © 2011-2022 走看看