笔试强训day10

//枚举两个端点

class Solution {
    bool check(int i, int j, string& s)
    {
        string t;
        for (int k = j; k >= i; --k)
        {
            t += s[k];
        }
        return t == s.substr(i,j-i+1);
    }
public:
    int getLongestPalindrome(string A) {
        // write code here
        int n = A.size(), len = 0;
        for (int i = 0; i < n; ++i)
        {
            for (int j = i; j < n; ++j)
            {
                if (check(i, j, A))
                    len = max(len, j - i + 1);
            }
        }
        return len;
    }
};

//枚举中心的点（中心扩展算法）
class Solution {
  public:
    int getLongestPalindrome(string s) {
        // write code here
        int n = s.size();
        int ret = 1;
        for (int i = 0; i < n; ++i) {
            int left = i - 1, right = i + 1;
            while (left >= 0 && right < n && s[left] == s[right]) {
                --left, ++right;
            }
            ret = max(ret, right - left - 1);
            left = i, right = i + 1;
            while (left >= 0 && right < n && s[left] == s[right]) {
                --left, ++right;
            }
            ret = max(ret, right - left - 1);
        }
        return ret;
    }
};

//动态规划
#include <iostream>
#include <vector>

using namespace std;
int n;
const int N = 100010;
int price[N];

int main() {
    cin >> n;
    for (int i = 0; i < n; i++) {
        scanf("%d", &price[i]);
    }
    vector<vector<int>>f(n, vector<int>(2));
    auto g = f;
    //f[i][j]代表的是处于买入状态的最大的利润
    //g[i][j]代表的是处于可交易状态的最大的利润
    f[0][0] = -price[0];
    for (int i = 1; i < n; i++) {
        for (int j = 0; j < 2; j++) {
            f[i][j] = max(f[i - 1][j], g[i - 1][j] - price[i]);
            g[i][j] = g[i - 1][j];
            if (j >= 1)g[i][j] = max(g[i][j], f[i - 1][j - 1] + price[i]);
        }
    }
    printf("%d\n", max(g[n - 1][0], g[n - 1][1]));
    return 0;
}

//贪心
#include <iostream>
#define inf 0x3f3f3f3f

using namespace std;

int main() {
    int n;
    cin>>n;
    int mi = inf,ma=-inf;
    int ans = 0;
    while(n--)
    {
        int x;
        cin>>x;
        mi = min(mi,x);
        ans = max(ans,x-mi);
    }
    cout<<ans<<'\n';
    return 0;
}
// 64 位输出请用 printf("%lld")

#include <iostream>
using namespace std;
int n, m, x, y;
long long dp[25][25];
int main() {
    cin >> n >> m >> x >> y;
    x += 1;
    y += 1;
    dp[0][1] = 1;
    for (int i = 1; i <= n + 1; i++) {
        for (int j = 1; j <= m + 1; j++) {
            if (i!=x&&j!=y&&abs(i - x) + abs(j - y) == 3 || (i == x &&
                    j == y)) {
                dp[i][j] = 0;
            } else {
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
    }
    cout << dp[n + 1][m + 1] << endl;
    return 0;
}