题目链接：https://ac.nowcoder.com/acm/contest/11255/G

题目大意：给你\(a,b,c\)，求满足\(lcm(a+x,b+y)=c\)的最小的\(x+y\)值

题目思路：设\(A=a+x,B=b+y\)，给的\(n\)和\(k_i\)很小，最多只有18，可以dfs暴力枚举所有符合条件的\(A\)和\(B\)，通过剪枝优化可以AC，具体看代码吧，另外这个题数字很大，需要用int128输入输出

思路和代码我参考的这位大佬nagisa_菜鸡

AC代码：

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <string>
#include <stack>
#include <deque>
#include <queue>
#include <cmath>
#include <map>
#include <set>
using namespace std;
template <typename T>
void write(T x)
{
    if (x < 0)
    {
        putchar('-');
        x = -x;
    }
    if (x > 9)
        write(x / 10);
    putchar(x % 10 + '0');
}
namespace FastIO
{
    const int SIZE = 1 << 16;
    char buf[SIZE], str[64];
    int l = SIZE, r = SIZE;
    int read(char *s)
    {
        while (r)
        {
            for (; l < r && buf[l] <= ' '; l++)
                ;
            if (l < r)
                break;
            l = 0, r = int(fread(buf, 1, SIZE, stdin));
        }
        int cur = 0;
        while (r)
        {
            for (; l < r && buf[l] > ' '; l++)
                s[cur++] = buf[l];
            if (l < r)
                break;
            l = 0, r = int(fread(buf, 1, SIZE, stdin));
        }
        s[cur] = '\0';
        return cur;
    }
    template <typename type>
    bool read(type &x, int len = 0, int cur = 0, bool flag = false)
    {
        if (!(len = read(str)))
            return false;
        if (str[cur] == '-')
            flag = true, cur++;
        for (x = 0; cur < len; cur++)
            x = x * 10 + str[cur] - '0';
        if (flag)
            x = -x;
        return true;
    }
    template <typename type>
    type read(int len = 0, int cur = 0, bool flag = false, type x = 0)
    {
        if (!(len = read(str)))
            return false;
        if (str[cur] == '-')
            flag = true, cur++;
        for (x = 0; cur < len; cur++)
            x = x * 10 + str[cur] - '0';
        return flag ? -x : x;
    }
} // namespace FastIO
using FastIO::read;
typedef __int128 int128;
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const int N = 110, M = 1e5;
const int base = 1e9;
const int P = 131;
const int mod = 998244353;
const double eps = 1e-6;
int n;
int128 a, b, c;
int128 pk[N];
int128 p[N], k[N];
int128 remain[N];
int128 ans;
void dfs(int u, int128 A, int128 B)
{
    if (A * remain[u] < a || B * remain[u] < b) //当前值和剩余所有的因数的乘积不够大
        return;
    if (A - a + B - b > ans) //当前求出的值不满足最优
        return;
    if (u > n)
    {
        ans = min(ans, A - a + B - b);
        return;
    }
    int128 pi = 1;
    for (int i = 1; i <= k[u]; ++i)
    {
        dfs(u + 1, A * pi, B * pk[u]);
        dfs(u + 1, A * pk[u], B * pi);
        pi *= p[u];
    }
    dfs(u + 1, A * pk[u], B * pk[u]); //减少一次dfs
}
int main()
{
    read(n);
    c = 1;
    for (int i = 1; i <= n; ++i)
    {
        read(p[i]), read(k[i]);
        pk[i] = 1;
        for (int j = 1; j <= k[i]; ++j)
            pk[i] *= p[i];
        c *= pk[i];
    }
    remain[n + 1] = 1;
    for (int i = n; i >= 1; --i)
        remain[i] = remain[i + 1] * pk[i];
    read(a), read(b);
    ans = 2 * c;
    dfs(1, 1, 1);
    write(ans);
    return 0;
}