f[i,j] = MAX(f[i-1,j-1]+a[i,j],f[i-1,j]+a[i,j])

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 510, INF = 1e9;

int n;
int a[N][N];
int f[N][N];

int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= i; j ++ )
            scanf("%d", &a[i][j]);

    for (int i = 0; i <= n; i ++ )
        for (int j = 0; j <= i + 1; j ++ )//注意這里j從0到i+1,因?yàn)閷?duì)于邊界點(diǎn),它的上一層只有一條路徑通向它
            f[i][j] = -INF;//初始化近似為-∞

    f[1][1] = a[1][1];
    for (int i = 2; i <= n; i ++ )
        for (int j = 1; j <= i; j ++ )
            f[i][j] = max(f[i - 1][j - 1] + a[i][j], f[i - 1][j] + a[i][j]);

    int res = -INF;
    for (int i = 1; i <= n; i ++ ) res = max(res, f[n][i]);

    printf("%d\n", res);
    return 0;
}

#include <iostream>

using namespace std;

const int N = 1010;

int n;
int w[N], f[N];

int main() {
    cin >> n;
    for (int i = 0; i < n; i++) cin >> w[i];

    int mx = 1;    // 找出所計(jì)算的f[i]之中的最大值，邊算邊找
    for (int i = 0; i < n; i++) {
        f[i] = 1;    // 設(shè)f[i]默認(rèn)為1，找不到前面數(shù)字小于自己的時(shí)候就為1
        for (int j = 0; j < i; j++) {
            if (w[i] > w[j]) f[i] = max(f[i], f[j] + 1);    // 前一個(gè)小于自己的數(shù)結(jié)尾的最大上升子序列加上自己，即+1
        }
        mx = max(mx, f[i]);
    }

    cout << mx << endl;
    return 0;
}

#include <iostream>

using namespace std;

const int N = 1010;
int n, cnt;
int w[N], f[N];

int main() {
    cin >> n;
    for (int i = 0 ; i < n; i++) cin >> w[i];

    f[cnt++] = w[0];
    for (int i = 1; i < n; i++) {
        if (w[i] > f[cnt-1]) f[cnt++] = w[i];
        else {
            int l = 0, r = cnt - 1;
            while (l < r) {
                int mid = (l + r) >> 1;
                if (f[mid] >= w[i]) r = mid;
                else l = mid + 1;
            }
            f[r] = w[i];
        }
    }
    cout << cnt << endl;
    return 0;
}

#include <iostream>

using namespace std;

const int N = 1010;

int n , m;
char a[N] , b[N];
int f[N][N];
int main()
{
    cin >> n >> m;
    cin >> a + 1 >> b + 1;

    for(int i = 1 ; i <= n ; i++)
        for(int j = 1 ; j <= m ; j++)
            {
                f[i][j] = max(f[i - 1][j] , f[i][j - 1]);//2和3的情況一定存在，所以可以無條件優(yōu)先判斷
                if(a[i] == b[j]) f[i][j] = max(f[i][j] , f[i - 1][j - 1] + 1);
            }                                                       

    cout << f[n][m] << endl;
    return 0;
}