【codeforces】gym 101138 K. The World of Trains【前缀和优化dp】

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题目链接：K. The World of Trains

记录一个横着的前缀dp和以及斜着的前缀dp，复杂度$O(n^2)$

#include <bits/stdc++.h>
using namespace std ;typedef pair < int , int > pii ;
typedef long long LL ;#define clr( a , x ) memset ( a , x , sizeof a )const int MAXN = 3500 ;
const int mod = 1e9 + 7 ;int dp[MAXN][MAXN] , f[MAXN][MAXN] , g[MAXN][MAXN] ;
int n , d , t , k ;int pm ( int x , int n ) {LL res = 1 ;while ( n ) {if ( n & 1 ) res = res * x % mod ;x = 1LL * x * x % mod ;n >>= 1 ;}return res ;
}void solve () {if ( k == 1 ) {if ( t + d - 1 == n ) printf ( "1\n" ) ;else printf ( "0\n" ) ;return ;}clr ( dp , 0 ) ;clr ( f , 0 ) ;clr ( g , 0 ) ;dp[0][0] = 1LL * k * pm ( k - 1 , mod - 2 ) % mod ;f[0][0] = 0 ;for ( int i = 1 ; i <= n ; ++ i ) {for ( int j = 0 ; j <= t ; ++ j ) {f[i][j] = ( f[i - 1][j] + dp[i - 1][j] ) % mod ;if ( i - d >= 0 ) f[i][j] = ( f[i][j] - dp[i - d][j] + mod ) % mod ;dp[i][j] = ( dp[i][j] + 1LL * ( k - 1 ) * f[i][j] ) % mod ;if ( i - d >= 0 && j ) {g[i][j] = ( g[i - 1][j - 1] + dp[i - d][j - 1] ) % mod ;}dp[i][j] = ( dp[i][j] + 1LL * ( k - 1 ) * g[i][j] ) % mod ;}}printf ( "%d\n" , dp[n][t] ) ;
}int main () {while ( ~scanf ( "%d%d%d%d" , &n , &d , &t , &k ) ) solve () ;return 0 ;
}

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