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  • 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;
    }

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  • 原文地址:https://www.cnblogs.com/jhz033/p/5755075.html
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