小a的子序列 (线性dp)
思路:设dp[i][j]表示最大数为j,i为第i的位置的萌值。那么推导过程就是两种情况:1.第i位数不放数字,则结果就是dp[i-1][j]; 2.第i位放数字,则结果就是前面的萌值sum+dp[i-1][j]*j
#include<iostream> using namespace std; #define ll long long const int maxn = 5050; const int mod = 1e9 + 7; ll dp[maxn][maxn]; int n, v; int main(){ cin >> n >> v; for (int i = 1; i <= v; ++i)dp[1][i] = 1; for (int i = 2; i <= n; ++i){ ll sum = 1; for (int j = 1; j <= v; ++j){ dp[i][j] = (dp[i - 1][j] + sum) % mod; sum = (sum + dp[i - 1][j]*j) % mod; } } ll ans = 0; for (int i = 1; i <= v; ++i)ans = (ans + dp[n][i]) % mod; cout << ans << endl; }