SGU 106. The equation 解题报告（模线性方程）

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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 -30 40 4

Sample Output

解题报告：简单的题意。方程ax+by+c=0在矩形((x1, y1), (x2, y2))中有多少个整点。

首先，应该考虑特殊情况，a=0，或者b=0，或者同时等于0。剩下的情况必定是一条有斜率K的直线了。

我的做法是在每个限制(x1, x2, y1, y2)中找最近的符合条件的点，按x排序，最大减最小，除以x的点距即可。

做法不难，但是WA了很多次，都是一些细节问题。如果真的是比赛，是要吃大亏的……码力仍需加强。

代码如下：

#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>using namespace std;typedef long long LL;// 扩展欧几里得void exgcd(LL a, LL b, LL &d, LL &x, LL &y){    if(b==0)        d=a, x=1, y=0;    else        exgcd(b, a%b, d, y, x), y-=x*(a/b);}// 解线性方程ax+by=c，返回最小的非负x解LL linear_mod(LL a, LL b, LL c){    LL d, x, y;    exgcd(a, b, d, x, y);   // 解ax+by=d, d为a，b最大公约数    if(c%d) return -1;      // 如果d不能整除c，方程无解    a/=d, b/=d, c/=d;       // 都除以最大公约数    if(b<0) b=-b;           // 取正b，如果使用y则y也要换号，此处忽略y    return (x*c%b+b)%b; // 该方程最小非负x解}LL limit_linear_mod(LL a, LL b, LL c, LL limit){    return linear_mod(a, b, c-a*limit)+limit;}LL flip_limit_linear_mod(LL a, LL b, LL c, LL limit){    return -limit_linear_mod(a, b, -c, -limit);}inline bool check(int x, int sta, int end){    return sta<=x && x<=end;}inline LL absLL(LL a){    return a<0?-a:a;}struct Point{    LL x, y;    bool operator<(const Point& cmp) const    {        return x<cmp.x;    }} point[10];int pointNum;LL xx1, xx2, yy1, yy2;void addPoint(LL x, LL y){    if(check(x, xx1, xx2) && check(y, yy1, yy2))        point[pointNum].x=x, point[pointNum].y=y, pointNum++;}LL gcd(LL a, LL b){    return b==0?a:gcd(b, a%b);}int main(){    LL a, b, c;    while(~scanf("%lld%lld%lld", &a, &b, &c))    {        scanf("%lld%lld%lld%lld", &xx1, &xx2, &yy1, &yy2);        c=-c;        LL ans=0;        if(a==0 && b==0)        {            if(c==0)                ans=(xx2-xx1+1)*(yy2-yy1+1);            printf("%lld\n", ans);            continue;        }        if(b==0)        {            swap(a, b);            swap(xx1, yy1);            swap(xx2, yy2);        }        if(a==0)        {            if(c%b==0 && check(c/b, yy1, yy2))                ans=(xx2-xx1+1);        }        else        {            LL tmpx, tmpy;            pointNum=0;            tmpx = limit_linear_mod(a, b, c, xx1);            tmpy = (c-a*tmpx)/b;            addPoint(tmpx, tmpy);            tmpy = limit_linear_mod(b, a, c, yy1);            tmpx = (c-b*tmpy)/a;            addPoint(tmpx, tmpy);            tmpx = flip_limit_linear_mod(a, b, c, xx2);            tmpy = (c-a*tmpx)/b;            addPoint(tmpx, tmpy);            tmpy = flip_limit_linear_mod(b, a, c, yy2);            tmpx = (c-b*tmpy)/a;            addPoint(tmpx, tmpy);            if(pointNum)            {                sort(point, point+pointNum);                ans=(absLL(point[pointNum-1].x-point[0].x)/absLL(b/gcd(a, b))+1);            }        }        printf("%lld\n", ans);    }}

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