2021 Niu Guest Winter Holiday Algorithm Basic Training Camp 1 A-string (dp push status)

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Topic

• How many strings consist of lowercase letters that are not more than n and contain the subsequence "us"? The answer is modulo 1e9+7.
• Subsequence, us may not be adjacent
• Range: 2≤n≤1e6

Ideas

• Use dp to transfer state. If you look at the mathematical formula to push the point here -
• Definition: dp[i][0] means a string of length i does not have u, dp[i][1] means a string of length i has u but no s, dp[i][2] means a string of length i There are us subsequences.
• Initialization: dp[1][0] = 25, dp[1][1] = 1, dp[1][2] = 0.
• State transition: dp[i][0] = dp[i-1][0] * 25 (there is no u before, and u is not selected in the current bit)
•                   dp[i][1] = dp[i-1][0] (there is no u before, the current position is u) + dp[i-1][1]*25 (u has been selected before but no s, currently Do not choose s) 
•                   dp[i][2] = dp[i-1][1] (previously there is u but no s, the current position is s) + dp[i-1][2] * 26 (there is already us before, so currently Choose any)
• The length does not exceed n, so dp[i][2] all meet the requirements

ac code

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int mod = 1e9 + 7;
const int maxn = 1e6 +  5;
ll dp[maxn][5];
int main(){
    int n; cin >> n;
    dp[1][0] = 25; //无u
    dp[1][1] = 1;  //有u无s
    dp[1][2] = 0;  //有us
    ll ans = 0;
    for(int i = 2; i <= n; i ++){
        dp[i][0] = dp[i - 1][0] * 25 % mod;
        dp[i][1] = (dp[i - 1][0] + dp[i - 1][1] * 25 % mod) % mod;
        dp[i][2] = (dp[i - 1][1] + dp[i - 1][2] * 26 % mod) % mod;
        ans = (ans + dp[i][2]) % mod;
    }
    cout << ans << endl;
    return 0;
}