Codeforces Round #455 (Div. 2) C. Python Indentation DP Recursion

Codeforces Round #455 (Div. 2)

C. Python Indentation

Test instructions: In Python, given n for loops or statements, ' F ' must have a statement. Combine the python indentation into a legitimate program and ask how many possible scenarios.

Tags:dp

DP[I][J] Indicates the number of possible scenarios at which the I-statement indent is J, and the transfer:

1 "If the first I is ' f ', then the first i+1 must be indented one unit more than the first, i.e. dp[i+1][j] = Dp[i][j].

2 "If the first I is ' s ', then the first i+1 can belong to any of the preceding for loop, that is, the indentation of the i+1 to <= the indent of the first I, that is dp[i+1][j] = Dp[i][k], j<=k<=n.

Complexity O (n^2).

#include<bits/stdc++.h> using namespace std; #pragma comment(linker, "/STACK:102400000,102400000") #define rep(i,a,b) for (int i=a; i<=b; ++i) #define per(i,b,a) for (int i=b; i>=a; --i) #define mes(a,b)  memset(a,b,sizeof(a)) #define INF 0x3f3f3f3f #define MP make_pair #define PB push_back #define fi  first #define se  second typedef long long ll; const int N = 5005, mod = 1e9+7; int n;
ll  dp[N][N]; char ch; int main()
{
    scanf("%d", &n);
    dp[1][0] = 1;
    rep(j,1,n) dp[1][j]=0;
    rep(i,1,n-1)
    {
        scanf("%*c%c", &ch); if(ch==‘f‘)
        {
            dp[i+1][0] = 0;
            rep(j,0,n) {
                dp[i+1][j+1] = dp[i][j];
            }
        } else {
            ll  sum = 0;
            per(j,n,0) {
                sum += dp[i][j];
                sum %= mod;
                dp[i+1][j] = sum;
            }

        }
    }
    scanf("%*c%c", &ch);
    ll  ans = 0;
    rep(j,0,n)
        ans += dp[n][j],  ans %= mod;
    printf("%lld\n", ans); return 0;
}

Codeforces Round #455 (Div. 2) C. Python Indentation DP recursion