请问这个怎么改能够让他不Time limit exceeded

Question

请问这个怎么改能够让他不Time limit exceeded

#include<stdio.h>
int answer(char a[], char b[], int m, int n) {
    int i, j, k, number = 0;
    if (n == 1) {
        for (i = 0; i < m; i++) {
            if (b[0] == a[i]) {
                number++;
            }
        }
        return number;
    }
    else if (n == 2) {
        i = 0;
        for (; i < m;i++)
        {
            if (b[0] == a[i]) {
                j = i + 1;
                for (; j < m; j++) {
                    if (b[1] == a[j]) {
                        number++;
                    }
                }
            }
        }return number;
    }
    else {
        i = 0;
        for (; i < m; i++)
        {
            if (b[0] == a[i]) {
                j = i + 1;
                for (; j < m; j++) {
                    if (b[1] == a[j]) {
                        k = j + 1;
                        for (; k < m; k++) {
                            if (b[2] == a[k]) {
                                number++; 
                            }
                        }
                    }
                }
            }
        }return number;
    }
}
int main() {
    int m, n, sum = 0; char ch;
    char a[100000], b[5];
    scanf("%d %d", &m, &n);
    scanf("%s",a);
    ch=getchar();
    scanf("%s", b);
    sum=answer(a,b,m,n);
    printf("%d\n", sum);
    return 0;
}

Answer 1

#include<stdio.h>
    int answer(char s[], char t[], int m, int n) {
        if(n > m)
            return 0;
    
        int dp[n + 1][m + 1];
        int len0 = m + 1;
        for (int i = 0; i < len0; i++) {
            dp[0][i] = 1;
        }
        int len = n + 1;
        for (int i = 1; i < len; i++) {
            dp[i][0] = 0;
        }

        for (int i = 1; i < len; i++) {
            for (int j = 1; j < len0; j++) {
                if(s[j - 1] == t[i - 1])
                    dp[i][j] = dp[i - 1][j - 1] + dp[i][j - 1];
                else
                    dp[i][j] = dp[i][j - 1];
            }
        }
        return dp[n][m];
    }
int main() {
    int m, n, sum = 0; char ch;
    char a[100000], b[5];
    scanf("%d %d", &m, &n);
    scanf("%s",a);
    ch=getchar();
    scanf("%s", b);
    sum=answer(a,b,m,n);
    printf("%d\n", sum);
    return 0;
}