歐幾里得演算法：

#include <cstdio>
#include <cstdlib>

/*
*   挑戰。。。p113
*/

struct point{   //格點
    int x;
    int y;
};

point p1, p2;   //兩個格點

int gcd(int a,int b){   //歐幾里得演算法
    if(b == 0) {
        return a;
    } else {
        return gcd(b, a % b);
    }
}

void solve(){
    int px = abs(p1.x - p2.x);  //兩格點的x座標距離
    int py = abs(p1.y - p2.y);  //兩格點的y左邊距離
    int res = 0;
    if(px != 0 && py != 0){
        res = gcd(px, py) - 1;
    }
    printf("%d\n", res);
}


int main(){
    while(scanf("%d%d%d%d", &p1.x, &p1.y, &p2.x, &p2.y) != EOF){
        solve();
    }
    return 0;
}
 

擴充套件歐幾里得演算法：

#include <cstdio>

/*
*   挑戰程式設計競賽 p115
/*

//思路：
//擴充套件歐幾里得演算法是在歐幾里得的演算法基礎上求解模線性方程及方程組
//已知ax + by = 1,gcd(a, b) = 1
//所以ax + by = gcd(a, b) ①
//再輾轉相除一次後，①變為②
//bx + (a % b)y = gcd(a, b) ②
//a % b = a - (a / b) * b ③
//將③代入②得到④
//bx + (a - (a / b) * b)y = gcd(a, b) ④
//將④化簡為⑤
//ay + b(x - (a / b) * y) = gcd(a, b) ⑤

//當b = 0時,顯然有 a = gcd(a , b),這時 x = 1, y = 0
//總之，有如下幾個等式
// 當 b = 0 時，x = 1, y = 0
// 其他情況 x = y, y = (x - (a / b) * y)

int a,b;
int x,y;
int extgcd(int a, int b, int &x, int &y){ //擴充套件歐幾里得演算法
    if(b == 0){
        x = 1;
        y = 0;
        return a;
    } else {
        int r = extgcd(b, a % b, x, y);
        int t = x;
        x = y;
        y = t - (a / b) * y;
        return r;
    }
}

void solve(){
    int res = extgcd(a, b, x, y);
    if(res != 1){
        printf("-1\n");
        return;
    }
    if(x > 0){
        printf("%d0", x);
    } else {
        printf("0%d", -x);
    }
    if(y > 0){
        printf("%d0\n", y);
    } else {
        printf("0%d\n", -y);
    }
}

int main(){
    while(scanf("%d%d", &a, &b) != EOF){
        solve();
    }
}