第二次测试部分题解

A——暴力枚举计数就好了，可以参考这段代码

#include <bits/stdc++.h>
using namespace std;
#define ll  long long
#define mod 100003
#define MAN 1000100
char str[10];
int ans;
void ok(){
	ans = 0;
	int len = 0;
	for(int i = 0; str[i] != '\0'; i++)
		len++;
	if(len == 1){
		ans = 0;
		for(char i = 'a'; i <= 'z'; i++){
			ans++;
			if(i == str[0])
				return;
		}
	}
	if(len == 2){
		ans = 26;
		for(char i = 'a'; i <= 'z'; i++){
			for(char j = i + 1; j <= 'z'; j++){
				ans++;
				if(str[0] == i && str[1] == j)
					return;
			}
		}
	}
	if(len == 3){
		ans = 351;
		for(char i = 'a'; i <= 'z'; i++){
			for(char j = i + 1; j <= 'z'; j++){
				for(char k = j + 1; k <= 'z'; k++){
					ans++;
					if(str[0] == i && str[1] == j && str[2] == k)
						return;
				}
			}
		}
	}
	if(len == 4){
		ans = 2951;
		for(char i = 'a'; i <= 'z'; i++){
			for(char j = i + 1; j <= 'z'; j++){
				for(char k = j + 1; k <= 'z'; k++){
					for(char l = k + 1; l <= 'z'; l++){
						ans++;
						if(str[0] == i && str[1] == j && str[2] == k && str[3] == l)
							return;
					}
				}
			}
		}
	}
	if(len == 5){
		ans = 17901;
		for(char i = 'a'; i <= 'z'; i++){
			for(char j = i + 1; j <= 'z'; j++){
				for(char k = j + 1; k <= 'z'; k++){
					for(char l = k + 1; l <= 'z'; l++){
						for(char p = l + 1; p <= 'z'; p++){
							ans++;
							if(str[0] == i && str[1] == j && str[2] == k && str[3] == l && str[4] == p)
								return;
						}
					}
				}
			}
		}
	}
}
int main(){
	while(scanf("%s", str) != EOF){
		int x = 1;
		for(int i = 0; str[i + 1] != '\0'; i++){
			if(str[i + 1] <= str[i]){
				x = 0;
				printf("0\n");
			}
		}
		if(x){
			ok();
			cout<<ans<<endl;
		}
	}
	return 0;
}

E—— 统计子矩阵

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

const int N = 510;
typedef long long LL;

int n, m, k;
int s[N][N];

int main() {
    cin >> n >> m >> k;
    for (int i = 1; i <= n; i ++)
        for (int j = 1; j <= m; j ++) {
            cin >> s[i][j];
            s[i][j] += s[i - 1][j]; // 每列上的一维前缀和
        }

    LL ans = 0;
    for (int i = 1; i <= n; i ++) // i j 上下边界
        for (int j = i; j <= n; j ++)
            for (int l = 1, r = 1, sum = 0; r <= m; r ++) { // l r 左右边界
                sum += s[j][r] - s[i - 1][r];
                while (sum > k) {
                    sum -= s[j][l] - s[i - 1][l];
                    l ++;
                }
                ans += r - l + 1;
            }

    cout << ans << endl;        
    
    return 0;
}

H—— Prefix Sum Addicts

#include <bits/stdc++.h>
 
using namespace std;
 
void solve()
{
	int n, k;
	cin >> n >> k;
	vector<long long> s(n + 1);
	for (int i = n - k + 1; i <= n; ++i)
		cin >> s[i];
	if (k == 1)
	{
		cout << "Yes" << endl;
		return;
	}
	vector<long long> a(n + 1);
	for (int i = n - k + 2; i <= n; ++i)
		a[i] = s[i] - s[i - 1];
	
	for(int i=n-k+3;i<=n;++i){
	    if (a[i-1]>a[i]){
    		cout << "No" << endl;
    		return;
	    }
	}
	
	if (s[n - k + 1] > a[n - k + 2] * (n - k + 1))
	{
		cout << "No" << endl;
		return;
	}
	cout << "Yes" << endl;
}
 
int main()
{
	int tests;
	cin >> tests;
	while (tests--) solve();
	return 0;
}

I——龟兔赛跑预测

#include <iostream>

using namespace std;

int v1, v2, t, s, l;

int main()
{
	scanf("%d %d %d %d %d", &v1, &v2, &s, &t, &l);
	int s1 = 0, s2 = 0,i=0;
	int t2 = l / v2;

	while(s1<l && s2<l)
	{
		if (s1 - s2 >= s)
		{
			s2 += (t * v2);
			i += t;
			if (i >= t2) break;
		}
		else
		{
			s1 += v1;
			s2 += v2;
			i++;
		}
	}
	if (s1 == s2 ) printf("D\n%d", t2);
	else if (s1 > s2 ) printf("R\n%d", i);
	else if (s1 < s2 ) printf("T\n%d", t2);

	return 0;
}