A. Stair, Peak, or Neither?（模拟）

#include<iostream>
using namespace std;
 
int main(){
    int t;
    scanf("%d", &t);
    int a, b, c;
    while(t--){
        scanf("%d%d%d", &a, &b, &c);
        if(a < b && b < c) printf("STAIR\n");
        else if(a < b && b > c) printf("PEAK\n");
        else printf("NONE\n");
    }
    return 0;
}

B. Upscaling（模拟）

#include <iostream>
using namespace std;
char g[100][100];
 
int main(){
    int t;
    scanf("%d", &t);
    while(t--){
        int n;
        scanf("%d", &n);
        int f = 0;
        
        for(int i = 1; i <= 2 * n; i++){
            for(int j = 1; j <= 2 * n; j += 2){
                if(i % 2 == 0){
                    cout << g[i - 1][j] << g[i - 1][j];
                } else {
                    if(f == 0){
                        if(g[i - 2][j] == '#'){
                            g[i][j] = '.';
                            g[i][j + 1] = '.';
                            cout << "..";
                        } else {
                            g[i][j] = '#';
                            g[i][j + 1] = '#';
                            cout << "##";
                            f = 1;
                        }
                    } else if(f == 1){
                        if(g[i - 2][j] == '.'){
                            g[i][j] = '#';
                            g[i][j + 1] = '#';
                            cout << "##";
                        } else {
                            g[i][j] = '.';
                            g[i][j + 1] = '.';
                            cout << "..";
                            f = 0;
                        }
                    }
                }
            }
            cout << endl;
        }
        //cout << endl;
    }
    
    return 0;
}

C. Clock Conversion（模拟）

#include <iostream>
#include <string>
 
using namespace std;
 
int main() {
    int t;
    cin >> t;
 
    for (int i = 0; i < t; i++) {
        string s;
        cin >> s;
 
        int hours = stoi(s.substr(0, 2));
        int minutes = stoi(s.substr(3, 2));
 
        if (hours >= 12) {
            if (hours > 12) {
                hours -= 12;
            }
            cout << (hours < 10 ? "0" : "") << hours << ":" << (minutes < 10 ? "0" : "") << minutes << " PM" << endl;
        } else {
            if (hours == 0) {
                hours = 12;
            }
            cout << (hours < 10 ? "0" : "") << hours << ":" << (minutes < 10 ? "0" : "") << minutes << " AM" << endl;
        }
    }
 
    return 0;
}

D. Product of Binary Decimals（dfs）

指数型枚举每一位是 0 还剩 1，然后除掉这个数，再继续搜索

#include<iostream>
using namespace std;
int a[20];
int f;
 
int check(int x) {
    while (x) {
        int digit = x % 10;
        if (digit > 1) return 0;
        x /= 10;
    }
    return 1;
}
 
void dfs(int u, int n) {
    if (u > 5) {
        int sum = 0;
        for (int i = 1; i <= 5; i++) {
            sum = sum * 10 + a[i];
        }
        if(sum == 0 || sum == 1) return;
        if (n % sum == 0){
            if(n / sum == 1){
                f = 1;
                return;
            }else{
                dfs(1, n / sum);
            }
        }
        return;
    }
    
    a[u] = 1;
    dfs(u + 1, n);
    a[u] = 0;
    
    a[u] = 0;
    dfs(u + 1, n);
    a[u] = 0;
}
 
int main() {
    int t;
    cin >> t;
    int n;
    while (t--) {
        f = 0;
        cin >> n;
        if(check(n)) f = 1;
        else dfs(1, n);
        if (f) cout << "YES" << endl;
        else cout << "NO" << endl;
    }
    return 0;
}

E. Nearly Shortest Repeating Substring（思维）

因为最多只能有一个字符不同，那么每个长度的区间只需要枚举连续的两段，判断即可

#include<iostream>
using namespace std;
int st[50];
 
int main(){
    int t, n;
    string s;
    cin>>t;
    while(t--){
        cin>>n>>s;
        int res = n;
        for(int len = 1; len <= n; len++){
            if(n % len != 0) continue;
            string m = s.substr(0, len);
            int cnt1 = 0, cnt2 = 0;
            for(int i = 0; i < n; i+=len){
                for(int j = 0; j < len; j++){
                    if(m[j] != s[i + j]) cnt1++;
                }
            }
            m = s.substr(len, len + len);
            for(int i = 0; i < n; i+=len){
                for(int j = 0; j < len; j++){
                    if(m[j] != s[i + j]) cnt2++;
                }
            }
            if(cnt1 <= 1 || cnt2 <= 1){
                res = len;
                break;
            }
        }
        cout<<res<<endl;
    }
    return 0;
}

F. 0, 1, 2, Tree!（二叉树）

#include<iostream>
using namespace std;

int main(){
	int t;
	cin>>t;
	int a, b, c;
	while(t--){
		cin>>a>>b>>c;
		if(c != a + 1) cout<<-1<<endl;
		else if(a + b + c == 1) cout<<0<<endl;
		else{
			// 存储当前父节点，子节点，树的高度 
			int cur = 1, ne = 0, res = 1;
			for(int i = 0; i < a + b; i++){
				// 如果父节点用完了，子节点就变成父节点 
				if(!cur){
					swap(cur, ne);
					res++;
				}
				cur--, ne++;
				// 如果两个节点的还没用完，那就先用它 
				if(i < a) ne++;
			}
			cout<<res<<endl; 
		}
	}
	return 0;
}