UVALive 6428 A+B // 扩展欧几里德

Problem Description

There is a computer, which has two memory cells (let us denote these cells by the letters a and b). Each cell (variable) stores some integer at any time. The computer can execute only two instructions a+=b and b+=a. The first instruction increases the value of the variable a by the value stored in the variable b. The second one, respectively, increases the value of b by the value a. A program for this computer consists of a sequence (possible empty) of such instructions. The instructions are executed in the appropriate order. Your task is to determine whether the given value S can be obtained in some cell after executing some program.

Input

The input file contains several test cases, each of them as described below.
The input contains three integers: the initial value of the variable a, the initial value of the variableb and the required value S (0 ≤ a, b, S ≤ 10^18).

Output

For each test case, write to the output ‘YES’ if the required value can be obtained as a result of some program execution, or ‘NO’ otherwise on a line by itself.

Sample Input

1 2 3
3 4 5
3 4 17

Sample Output

YES
NO
YES

解题思路

判断 $ax +by = s$ ……(1)是否存在非负整数解, 且满足 $gcd(x, y) = 1$。

根据扩展欧几里得算法知，存在整数(x, y) 满足 $ax + by = gcd(a, b)$;
因此我们可以先根据 【gcd(a, b) 能否整除 s】 来判断 是否存在 整数解。

引理：设$a, b, c$为任意整数。若方程$ax+by=c$的一组整数解为$({x}_{0}, {y}_{0})$,则它的任意整数解都可以写成$({x}_{0}+k*b', {y}_{0}-k*a')$, 其中$a' = a/gcd(a, b), b' = b/gcd(a, b)$

令 $d = gcd(a, b);$
(1)式 转化为 $a'x + b'y = s'$……(2)
其中 $a' = a/d, b' = b/d, s' = s/d;$
然后求出一组 $x, y$ 满足$a'x + b'y = 1$……(3);
这个算出的 $x, y$ 要乘上 $s'$才是和 (2)式等价 即$x *= s', y *= s';$
然后根据 $x$ 找到最小非负整数解 ${x}_{0}$ , 根据引理，只要 $x$ %= $b'$，若为负数，则加$b$至非负为止。
最后枚举所有$x，y$即可。

参考代码

#include <stdio.h>
typedef long long ll;
//扩展欧几里得算法 : 求解满足ax+by=1的一组解，且d = gcd(a, b);
void exgcd(ll a, ll b, ll& d, ll& x, ll& y) {
    if (!b) {
        d = a; x = 1; y = 0;
    } else {
        exgcd(b, a % b, d, y, x);
        y -= x * (a / b);
    }
}
ll gcd(ll a, ll b){return b ? gcd(b, a % b) : a;}
ll a, b, s, d, x, y;
void work() {
    // 特判有0的情况
    if (a == 0) {
        if (b == 0) printf("%s\n", s ? "NO" : "YES");
        else printf("%s\n", s % b ? "NO" : "YES");
        return ;
    } else if (b == 0) {
        printf("%s\n", s % a ? "NO" : "YES");
        return ;
    }
    exgcd(a, b, d, x, y);
    if (s % d) {
        puts("NO");
        return ;
    }
    a /= d; b /= d; s /= d;
    x *= s % b; x %= b;
    while (x < 0) x += b;
    y = (s - a * x) / b;
    while (y >= 0) {
        if (gcd(x, y) == 1) {
            puts("YES");
            return ;
        }
        x += b; y -= a;
    }
    puts("NO");
}
int main() {
    while (~scanf("%lld %lld %lld", &a, &b, &s))  work();
    return 0;
}