sgu 106. The equation 已知ax + by + c = 0 ，求解在[x1,x2],[y1,y2]区间的解的个数

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本文介绍了一种算法，用于解决给定线性方程 ax + by + c = 0 的情况下，在指定范围内找到所有满足条件的整数根 (x, y) 对的方法。通过扩展欧几里得算法计算最大公约数，并确定特定条件下整数根的数量。

106. The equation

time limit per test: 0.50 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

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
__int64 exgcd(__int64 a,__int64 b,__int64 &x,__int64 &y)
{
    if(b==0)
    {
        x=1,y=0;
        return a;
    }
    __int64 r=exgcd(b,a%b,x,y);
    __int64 t=x;x=y;y=t-a/b*y;
    return r;
}
int main()
{
    __int64 a,b,c,x1,x2,y1,y2,x,y;
    while(scanf("%I64d%I64d%I64d",&a,&b,&c)==3)
    {
        c=-c;
        scanf("%I64d%I64d%I64d%I64d",&x1,&x2,&y1,&y2);
        if(x1 > x2 || y1 > y2)
        {
            printf("0/n");
            continue;
        }
        //特殊情况a和b有1个为0 和 a和b同时为0
        if(!a||!b)
        {
            if(!a&&!b&&!c)//a和b同时为0
            {
                __int64 res1 = x2 - x1 + 1,res2 = y2 - y1 + 1;//会溢出,注意精度
                res1 = res1 * res2;
                printf("%I64d/n", res1);
                continue;
            }
            else if(!a&&!b&&c) //a和b同时为0
            {
                printf("0/n");continue;
            }
            else if(!a&&b)//a和b有1个为0 a为0
            {
                if(c%b)
                {
                    printf("0/n");continue;
                }
                else
                {
                    y=c/b;
                    if(y>=y1&&y<=y2) printf("%I64d/n",x2-x1+1);
                    else printf("0/n");
                    continue;
                }
            }
            else if(a&&!b)//a和b有1个为0 b为0
            {
                if(c%a)
                {
                    printf("0/n");continue;
                }
                else
                {
                    x=c/a;
                    if(x>=x1&&x<=x2) printf("%I64d/n",y2-y1+1);
                    else printf("0/n");
                    continue;
                }
            }
        }
        __int64 r=exgcd(a,b,x,y);
        if(c%r)
        {
            printf("0/n");continue;
        }
        x*=c/r,y*=c/r;//算出一个解
        __int64 _count=0;
        __int64 lx, rx, ly, ry;
        //x=x0+k*b/r,y=y0-k*a/r;
        //* *r/b== * /(b/r),* *r/a== * /(b/a)
        lx = (x1<=x || (x1-x)*r%b==0) ? (x1-x)*r/b : (x1-x)*r/b+1;
        rx = (x2>=x || (x-x2)*r%b==0) ? (x2-x)*r/b : (x2-x)*r/b-1;
        ly = (y1<=y || (y1-y)*r%a==0) ? (y-y1)*r/a : (y-y1)*r/a-1;
        ry = (y2>=y || (y-y2)*r%a==0) ? (y-y2)*r/a : (y-y2)*r/a+1;
        if (lx > rx) swap(lx, rx);
        if (ly > ry)  swap(ly, ry);
        if (lx <= ry && ly <= rx) { //求出两个区间交集的元素个数
            __int64 _max = (lx>ly) ? lx : ly;
            __int64  _min = (rx<ry) ? rx : ry;
            _count = _min-_max+1;
        }
        printf("%I64d/n",_count);
    }
    return 0;
}