[bzoj5027][exgcd][乱搞]数学题

Description

给出a,b,c,x1,x2,y1,y2，求满足ax+by+c=0，且x∈[x1,x2],y∈[y1,y2]的整数解有多少对？

Input

第一行包含7个整数，a,b,c,x1,x2,y1,y2，整数间用空格隔开。 a,b,c,x1,x2,y1,y2的绝对值不超过10^8。

Output

输出整数解有多少对？

Sample Input

1 1 -3 0 4 0 4

Sample Output

4

题解

真是失了智写的什么辣鸡..
ex随便搞出一组解
然后瞎移范围
欢快re
拍了一晚上没出错
发现
我对拍没出0的情况..
然后被A=0或者B=0卡了一晚上.

#include
#include
#include
#include
#include
#define LL long long
#define mx 200000000
using namespace std;
LL exgcd(LL a,LL b,LL &x,LL &y)
{
    if(a==0)
    {
        x=0;y=1;
        return b;
    }
    else
    {
        LL tx,ty;
        LL d=exgcd(b%a,a,tx,ty);
        x=ty-(b/a)*tx;
        y=tx;
        return d;
    }
}
LL A,B,C,K;
LL u1,u2,v1,v2;
int main()
{
    scanf("%lld%lld%lld%lld%lld%lld%lld",&A,&B,&C,&u1,&u2,&v1,&v2);
    if(u1>u2||v1>v2){printf("0\n");return 0;}
    if(!A&&!B)
    {
        if(C==0)printf("%lld\n",(LL)(u2-u1+1)*(v2-v1+1));
        else printf("0\n");
        return 0;
    }
    K=-C;LL x,y;
    if(A==0)
    {
        if(K%B){puts("0");return 0;}
        if(K/B>=v1&&K/B<=v2)printf("%lld\n",u2-u1+1);
        else puts("0");
        return 0;
    }
    if(B==0)
    {
        if(K%A){puts("0");return 0;}
        if(K/A>=u1&&K/A<=u2)printf("%lld\n",v2-v1+1);
        else puts("0");
        return 0;
    }
    LL d=exgcd(A,B,x,y);
    if(K%d!=0){printf("0\n");return 0;}
    x=(x*(K/d));
//  printf("%lld\n",(K-A*x)/B);
    if(x>u2)
    {
        LL gg=abs(B/d),dis=x-u1;//相邻可行x相距的长度
        if(gg==0){printf("0\n");return 0;}
        x=x-(dis/gg)*gg;
    }
    else if(xabs(B/d),dis=u2-x;
        if(gg==0){printf("0\n");return 0;}
        x=x+(ceil(double(dis/gg)))*gg;
        gg=abs(B/d),dis=x-u1;
        x=x-(dis/gg)*gg;
    }
    else
    {
        LL gg=abs(B/d),dis=x-u1;
        if(gg!=0)x=x-(dis/gg)*gg;
    }
    if(x>u2||xprintf("0\n");return 0;}
    y=(K-A*x)/B;
    LL ln1=abs(A/d),ln2=abs(B/d);//相邻可行y相距的长度 相邻可行x相距的长度
    LL u=x+ln2;
    LL v=(K-A*u)/B;
    int op;
    if(v<y)op=0;//y随着x增大而减小 
    else op=1;//增大而增大
    if(op==0)//减小 
    {
        if(yprintf("0\n");return 0;}
        if(y>v2)
        {
            if(ln1==0){printf("0\n");return 0;}
            LL ad1=ceil(double(y-v2)/double(ln1)),ad2=(u2-x)/ln2;
            if(ad1>ad2){printf("0\n");return 0;}
            y=y-ln1*ad1;x=x+ln2*ad1;
        }
        if(yy>v2){printf("0\n");return 0;}
        LL can_go=min((ln2==0?999999999:(u2-x)/ln2),(ln1==0?999999999:(y-v1)/ln1));
        printf("%lld\n",can_go+1);
    }
    else
    {
        if(y>v2){printf("0\n");return 0;}
        if(yif(ln1==0){printf("0\n");return 0;}
            LL ad1=(v2-y)/ln1,ad2=(u2-x)/ln2;
            y=y+ln1*ad1;
            LL gg1=(y-v1)/ln1;
            y=y-ln1*gg1;
            ad1-=gg1;
            if(ad1>ad2){printf("0\n");return 0;}
            x=x+ln2*ad1;
        }
        if(yy>v2){printf("0\n");return 0;}
        LL can_go=min((ln2==0?999999999:(u2-x)/ln2),(ln1==0?999999999:(v2-y)/ln1));
        printf("%lld\n",can_go+1);
    }
    return 0;
}