Codeforces 946F Fibonacci String Subsequences dp (看题解)

墨蓝 2021-12-17 00:17 200阅读 0赞

[Fibonacci String Subsequences][]

读题把子序列读成子串， 码都码完了， 样例不对， 人傻了。

我dp的是每个匹配串的贡献， 然后发现子序列的话算贡献维护不了。。

没想到直接dp的就是在子序列中出现的次数。

dp\[ i \]\[ l \]\[ r \] 表示的是在 F\[ i \]这个串中， s\[ l \] - s\[ r \]这个子串在所有子序列中的出现次数。

#include<bits/stdc++.h>
    #define LL long long
    #define LD long double
    #define ull unsigned long long
    #define fi first
    #define se second
    #define mk make_pair
    #define PLL pair<LL, LL>
    #define PLI pair<LL, int>
    #define PII pair<int, int>
    #define SZ(x) ((int)x.size())
    #define ALL(x) (x).begin(), (x).end()
    #define fio ios::sync_with_stdio(false); cin.tie(0);
    
    using namespace std;
    
    const int N = 100 + 7;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3f;
    const int mod = 1e9 + 7;
    const double eps = 1e-8;
    const double PI = acos(-1);
    
    template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
    template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
    template<class T, class S> inline bool chkmax(T& a, S b) {
        return a < b ? a = b, true : false;}
    template<class T, class S> inline bool chkmin(T& a, S b) {
        return a > b ? a = b, true : false;}
    
    LL power(LL a, LL b) {
        LL ans = 1;
        while(b) {
            if(b & 1) ans = ans * a % mod;
            a = a * a % mod; b >>= 1;
        }
        return ans;
    }
    
    int n, x;
    char s[N];
    LL fib[N];
    LL f[N][N][N];
    
    LL dp(int x, int l, int r) {
        LL& ans = f[x][l][r];
        if(~ans) return ans;
        ans = 0;
        if(x == 0) {
            if(l == r && s[l] == '0') ans = 1;
        }
        else if(x == 1) {
            if(l == r && s[l] == '1') ans = 1;
        }
        else {
            if(r == n) add(ans, dp(x - 1, l, r) * power(2, fib[x - 2]) % mod);
            else add(ans, dp(x - 1, l, r));
            if(l == 1) add(ans, dp(x - 2, l, r) * power(2, fib[x - 1]) % mod);
            else add(ans, dp(x - 2, l, r));
            for(int k = l; k < r; k++) add(ans, dp(x - 1, l, k) * dp(x - 2, k + 1, r) % mod);
        }
        return ans;
    }
    
    int main() {
        memset(f, -1, sizeof(f));
        fib[0] = fib[1] = 1;
        for(int i = 2; i <= 100; i++)
            fib[i] = (fib[i - 1] + fib[i - 2]) % (mod - 1);
        scanf("%d%d", &n, &x);
        scanf("%s", s + 1);
        printf("%lld", dp(x, 1, n));
        return 0;
    }
    
    /*
    */

转载于:https://www.cnblogs.com/CJLHY/p/10825050.html

[Fibonacci String Subsequences]: https://codeforces.com/problemset/problem/946/F