思路：求a mod 上b后的值为amodb, 求gcd(b, amodb)即可

int gcd(int a,int b){
	return b ? gcd(b, a % b) : a;
}

void solve(){
	string a;
	cin >> a;
	int b;
	cin >> b;
	int amodb = 0;
	for(auto c : a){
		amodb = (amodb * 10 + (c - '0')) % b;
	}
	cout << gcd(b, amodb);
}

思路：二分答案，然后，判断当前二分的值符不符合，如果可以走到(n,n)就缩小答案，否则扩大答案。

#include <bits/stdc++.h>
#define int long long
#define x first
#define y second
#define endl '\n'
#define pq priority_queue
using namespace std;
typedef pair<int, int> pii;

int n, a[550][550], st[550][550];
int dx[] = {-1, 1, 0, 0}, dy[] = {0, 0, -1, 1};

bool check(int mid) {
    queue<pii> q;
    memset(st, 0, sizeof(st));
    if (a[1][1] > mid) return false;
    q.push({1, 1});
    st[1][1] = true;
    
    while (!q.empty()) {
        int x = q.front().first, y = q.front().second;
        q.pop();
        
        for (int i = 0; i < 4; i++) {
            int px = x + dx[i], py = y + dy[i];
            if (px < 1 || py < 1 || px > n || py > n)
                continue;
            if (st[px][py])
                continue;
            if (a[px][py] > mid)
                continue;
            q.push({px, py});
            st[px][py] = true;
        }
    }
    return st[n][n];
}

void solve() {
    cin >> n;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            cin >> a[i][j];
        }
    }

    int l = 0, r = 1e9 + 2;
    while (l + 1 != r) {
        int mid = l + r >> 1;
        if (check(mid))
            r = mid;
        else
            l = mid;
    }
    cout << r << endl;
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    solve();
    return 0;
}