牛客326B 背单词

传送门：https://ac.nowcoder.com/acm/contest/326/B

题意：给你一个n，你需有找有多少个长度从1~n 的单词，满足最长连续元音少于A个并且最长连续辅音长度少于B。

题解：我们定义dp的状态是，dp1[i][j]表示长度为i的，最长连续元音长度为j的单词的个数，

dp2[i][j]表示长度为i的，最长连续辅音长度为j的单词的个数

由于元音的个数为5个可以任意取，辅音的个数为21个可以任意取

转移方程就是dp1[i][j]=(dp1[i][j]+dp1[i-1][j-1])*5,dp2[i][j]=(dp2[i][j]+dp2[i-1][j-1])*21

每次转移完记得更新dp1[i][1] 的状态和dp2[i][1]的状态，即前面全部是元音/辅音，但是后面变成辅音/元音时的个数

计数时记得取模

代码如下：

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <stack>
#include <queue>
#include <cstdio>
#include <cctype>
#include <bitset>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#define PI acos(-1)
#define eps 1e-8
#define fuck(x) cout<<#x<<" = "<<x<<endl;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w+",stdout);
//#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int, int> PII;
const int maxn = 3e5 + ;
const LL INF = 1e18 + ;
const ull mod = 1e9 + ;
LL gcd(LL a, LL b) {return b ? gcd(b, a % b) : a;}
LL lcm(LL a, LL b) {return a / gcd(a, b) * b;}
LL powmod(LL a, LL b, LL MOD) {LL ans = ; while (b) {if (b % )ans = ans * a % MOD; a = a * a % MOD; b /= ;} return ans;}
double dpow(double a, LL b) {double ans = 1.0; while (b) {if (b % )ans = ans * a; a = a * a; b /= ;} return ans;}
 
LL dp1[][];
LL dp2[][];
int main() {
#ifndef ONLINE_JUDGE
    FIN
#endif
    int T;
    scanf("%d", &T);
    while (T--) {
        memset(dp1,,sizeof(dp1));
        memset(dp2,,sizeof(dp2));
        int n, a, b;
        cin >> n >> a >> b;
        LL ans = ;
        dp1[][] = ;
        dp2[][] = ;
 
        for (int i = ; i <= n; i++) {
            LL sum1 = ;
            LL sum2 = ;
            for (int j = ; j <= min(a,i); j++) {
                dp1[i][j] = (dp1[i][j] + dp1[i - ][j - ] * ) % mod;
                sum1 = (sum1 + dp1[i - ][j]) % mod;
            }
            for (int j = ; j <= min(b,i); j++) {
                dp2[i][j] = (dp2[i][j] + dp2[i - ][j - ] * ) % mod;
                sum2 = (sum2 + dp2[i - ][j]) % mod;
            }
            dp1[i][] = (dp1[i][] + sum2 *  % mod) % mod;
            dp2[i][] = (dp2[i][] + sum1 *  % mod) % mod;
        }
        for (int i = ; i <= n; i++) {
            for (int j = ; j <= a; j++) {
                ans = (ans + dp1[i][j]) % mod;
            }
            for (int j = ; j <= b; j++) {
                ans = (ans + dp2[i][j]) % mod;
            }
        }
        cout << ans << endl;
    }
}