Codeforces Round #483 (Div. 2) [Thanks, Botan Investments and Victor Shaburov!]

A. Game

题解：排序后取中间的数。

B. Minesweeper

题解：检查每一位是否合法就行。

感受：永远都被B题坑.。。。。。if there is a digit $\mathrm{kk\; in\; the\; cell,\; then\; exactly\; kk\; neighboring\; cells\; have\; bombs.}$周围只是没炸弹而已，而不是全部'.'。WA了四次，强行掉分，囧rz。

#pragma warning(disable:4996)
#include<cmath>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define mem(arr,in) memset(arr,in,sizeof(arr))
using namespace std;

const int maxn = 1005;

int n, m;
char mp[200][200];

int dx[] = { -1,-1,0,1,1,1,0,-1 };
int dy[] = { 0,1,1,1,0,-1,-1,-1 };

bool check1(int a, int b) {
    for (int i = 0; i <= 7; i++) {
        int mx = dx[i] + a;
        int my = dy[i] + b;
        if (mx >= 0 && mx < n && my >= 0 && my < m && mp[mx][my] == '*') return false;
    }
    return true;
}

bool check2(int a, int b) {
    int cnt = 0;
    for (int i = 0; i <= 7; i++) {
        int mx = dx[i] + a;
        int my = dy[i] + b;
        if (mx >= 0 && mx < n && my >= 0 && my < m && mp[mx][my] == '*') cnt++;
    }
    int k = mp[a][b] - '0';
    if (cnt == k) return true;
    else return false;
}

bool Solve() {
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (mp[i][j] == '.') {
                if (!check1(i, j)) return false;
            }
            else if (mp[i][j] >= '1' && mp[i][j] <= '8') {
                if (!check2(i, j)) return false;
            }
        }
    }
    return true;
}

int main()
{
    while (cin >> n >> m) {
        for (int i = 0; i < n; i++) scanf("%s", mp[i]);
        //for (int i = 0; i < n; i++) printf("%s\n", mp[i]);
        if (Solve()) cout << "YES" << endl;
        else cout << "NO" << endl;
    }
    return 0;
}

C. Finite or not?

题解：while(((p*b)%q)) p=p*b%q;这样貌似会超时，比赛的时候不敢写。还有就是判断是否存在k使得，p*b^k/q=0，即q的所有素因子是否在b中存在。

注意：不要用long long!!!

#pragma warning(disable:4996)
#include<cmath>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long 
#define mem(arr,in) memset(arr,in,sizeof(arr))
using namespace std;

const int maxn = 1005;

int T;
ll p, q, b;

ll gcd(ll x, ll y) {
    if (y == 0) return x;
    return gcd(y, x % y);
}

int main()
{
    scanf("%d", &T);
    for (int i = 1; i <= T; i++) {
        scanf("%I64d %I64d %I64d", &p, &q, &b);
        if (p == 0) {
            printf("Finite\n");
        }
        else {
            ll d = gcd(p, q);
            q = q / d;
            while (1) {
                d = gcd(b, q);
                if (d == 1) break;
                while (q % d == 0) q = q / d;
            }
            if (q == 1) {
                printf("Finite\n");
            }
            else {
                printf("Infinite\n");
            }
        }
    }
    return 0;
}

 D. XOR-pyramid

题解：f ( l , r ) = f ( l , r - 1 ) ^ f ( l + 1 , r )，则 dp[ l ][ r ]=max(dp[ l ][ r ]，dp[ l ][ r - 1 ]，dp[ l + 1 ][ r ])。

感受：对位运算不熟啊，异或运算满足结合率和交换率。第一个式子，可以通过 f ( a , b , c , d)归纳出来，证明就不写了，第二个式子意思是，答案就在 f 递归中，要么是 f ( l , r ) , f ( l ,r - 1) , f( l + 1 , r)。如果不预处理出所有区间的最大值，那么就只有在询问中递归寻找最大值，算 f ( l , r )也就没意义了。

#pragma warning(disable:4996)
#include<cmath>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long 
#define mem(arr,in) memset(arr,in,sizeof(arr))
using namespace std;

const int maxn = 5005;

int n, m;
int dp[maxn][maxn];

int main()
{
    while (scanf("%d", &n) != EOF) {
        for (int i = 1; i <= n; i++) scanf("%d", &dp[i][i]);
        for (int i = 1; i <= n; i++) {
            for (int j = i - 1; j >= 1; j--) {
                dp[j][i] = dp[j][i - 1] ^ dp[j + 1][i];
            }
        }
        for (int i = 1; i <= n; i++) {
            for (int j = i - 1; j >= 1; j--) {
                dp[j][i] = max(dp[j][i], max(dp[j][i - 1], dp[j + 1][i]));
            }
        }
        scanf("%d", &m);
        for (int i = 1; i <= m; i++) {
            int l, r;
            scanf("%d%d", &l, &r);
            printf("%d\n", dp[l][r]);
        }
    }
    return 0;
}