A - Job Interview

#include <bits/stdc++.h>
using namespace std;

int main(){
	int n;
	string s;
	cin >> n >> s;
	if( s.find("x") != -1 ){
		printf("No\n");

	}else if( s.find("o") == -1 ){
		printf("No\n");
	}else printf("Yes\n");
	return 0;
}

B - Coloring Matrix

旋转三次然后暴力比较

#include <bits/stdc++.h>
using namespace std;

const int N = 105;

int a[N][N] , b[N][N] , c[N][N];

int main(){
	int n;
	cin >> n;
	for( int i = 1 ; i <= n ; i ++ )
		for( int j = 1 ; j <= n ; j ++ )
			cin >> a[i][j];
	for( int i = 1 ; i <= n ; i ++ )
		for( int j = 1 ; j <= n ; j ++ )
			cin >> b[i][j];
	for( int k = 4 ; k ; k -- ){
		bool f = true;
		for( int i = 1 ; f && i <= n ; i ++ ){
			for( int j = 1 ; f && j <= n ; j ++ ){
				if( a[i][j] == 1 && b[i][j] != 1 ) f = false; 
			}
		}
		if( f ) return printf("Yes\n") , 0;
		for( int i = 1 ; i <= n ; i ++ )
			for( int j = 1 ; j <= n ; j ++ )
				c[n+1-j][i] = a[i][j];
		for( int i = 1 ; i <= n ; i ++ )
			for( int j = 1 ; j <= n ; j ++ )
				a[i][j] = c[i][j];
	}
	printf("No\n");
	return 0;
}

C - Cards Query Problem

整两个set数组维护一下就好了，注意 box 中的 card 要用 multiset维护，因为会重复

#include <bits/stdc++.h>
using namespace std;

int read(){
	int ch = getchar() , x = 0;
	while( ch < '0' || ch > '9' ) ch = getchar();
	while( ch >= '0' && ch <= '9' ) x = (x<<3)+(x<<1) + ch - '0' , ch = getchar();
	return x;
}

const int N = 2e5+5;
int n;
multiset<int> box[N];
set<int> card[N];

int main(){
	int n = read() , q = read();
	for( int op , x , y ; q ; q -- ){
		op = read();
		if( op == 1 ){
			x = read() , y = read();
			box[y].insert(x) , card[x].insert(y);
		}else{
			x = read();
			if( op == 2 ){
				for( auto i : box[x] ) printf("%d " , i );
			}else{
				for( auto i : card[x] ) printf("%d " , i );
			}
			printf("\n");
		}
	}

}

D - Writing a Numeral

其实只要知道每个数被乘十了多少次就很好算了

#include <bits/stdc++.h>
using namespace std;

#define int long long

int read(){
	int ch = getchar() , x = 0;
	while( ch < '0' || ch > '9' ) ch = getchar();
	while( ch >= '0' && ch <= '9' ) x = (x<<3)+(x<<1) + ch - '0' , ch = getchar();
	return x;
}

const int mod = 998244353;

int power( int x , int y ){
	int ans = 1;
	while( y ){
		if( y & 1 ) ans = ans * x % mod;
		x = x * x % mod , y >>= 1;
	}
	return ans;
}

int32_t main(){
	vector<int> dq;
	int pos = 0;
	dq.push_back(1);
	int cnt = 1 , q = read();
	for( int op , x ; q ; q -- ){
		op = read();
		if( op == 1 ){
			x = read();
			cnt = (cnt * 10 % mod + x) % mod;
			dq.push_back(x);
		}else if( op == 2 ){
			x = dq[pos++];
			x = mod - x * power( 10 , dq.size() - pos ) % mod;
			cnt = ( cnt + x ) % mod;
		}else printf("%lld\n" , cnt);
	}

}

F - Rook Score

首先用map统计一下每一行每一列的和。然后考虑枚举(r,c)。这里可以分成两类情况。

第一种(r,c)位置上的值不是 0，此时枚举所有的点就好。

第二种(r,)位置上的值是 0 ，首先把所有的行列都取出来按照和从大到小排序。然后枚举行行列，如果找到了(r,c)不存在的情况就更新答案并 break，因为已经排序后面找到一定更小。

#include<bits/stdc++.h>

using namespace std;

#define int long long
typedef long long ll;

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

int32_t main() {
    int n = read();
    map<int, int> cntR, cntC;
    map<int, map<int, int>> val;
    set<pair<int, int>> st;
    for (int x, y, w; n; n--) {
        x = read(), y = read(), w = read();
        cntR[x] += w, cntC[y] += w, val[x][y] = w, st.emplace(x, y);
    }
    vector<pair<int, int>> r, c;
    for (auto [k, v]: cntR) r.emplace_back(v, k);
    for (auto [k, v]: cntC) c.emplace_back(v, k);
    sort(r.begin(), r.end(), greater<pair<int, int>>());
    sort(c.begin(), c.end(), greater<pair<int, int>>());
    int res = 0;
    for (auto [x, y]: st)
        res = max(res, cntR[x] + cntC[y] - val[x][y] );
    for( auto [ v1 , x ] : r )
        for( auto [ v2 , y ] : c ){
            if( st.count(make_pair(x,y) ) ) continue;
            res = max( res , v1 + v2 );
            break;
        }
    cout << res << "\n";
    return 0;
}