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【HDU】5291 Candy Distribution 【DP+打标记】

2015-07-27 15:44 441 查看
传送门:【HDU】5291 Candy Distribution

my code:my~~code:

[code]/*************************************************************************
    > File Name: D.cpp
    > Author: poursoul
    > Created Time: 2015年07月22日 星期三 13时04分24秒
 ************************************************************************/

#include <bits/stdc++.h>
using namespace std ;

typedef long long LL ;
typedef long long Int ;
typedef pair < int , int > pi ;

#define clr(a,x) memset ( a , x , sizeof a )

const int MAXN = 100005 ;
const int N = 50000 ;
const int mod = 1e9 + 7 ;

int dp[2][MAXN] ;
int c1[MAXN] , c2[MAXN] ;
int n ;

void update ( int& x , int y ) {
    x += y ;
    while ( x >= mod ) x -= mod ;
    while ( x < 0 ) x += mod ;
}

void solve () {
    int x , cur = 0 , sum = 0 ;
    scanf ( "%d" , &n ) ;
    clr ( dp[cur] , 0 ) ;
    dp[cur]
 = 1 ;
    for ( int i = 1 ; i <= n ; ++ i ) {
        cur ^= 1 ;
        scanf ( "%d" , &x ) ;
        int L1 = N - sum - x , R1 = N + sum + x ;
        int L2 = N - sum , R2 = N + sum ;
        sum += x ;
        for ( int j = L1 - 2 ; j <= R1 + 10 ; ++ j ) {
            c1[j] = c2[j] = dp[cur][j] = 0 ;
        }
        for ( int j = L2 ; j <= R2 ; ++ j ) if ( dp[cur ^ 1][j] ) {
            int s = dp[cur ^ 1][j] ;
            if ( x % 2 == 0 ) {
                update ( c1[j - x] , s ) ;
                update ( c1[j + 2] , -s ) ;
                update ( c2[j + 1] , -s ) ;
                update ( c2[j + x + 3] , s ) ;
            } else {
                update ( c1[j - x] , s ) ;
                update ( c1[j + 1] , -s ) ;
                update ( c2[j + 2] , -s ) ;
                update ( c2[j + x + 3] , s ) ;
            }
        }
        for ( int j = L1 ; j <= R1 ; ++ j ) {
            update ( c1[j] , c1[j - 2] ) ;
            update ( c2[j] , c2[j - 2] ) ;
        }
        for ( int j = L1 ; j <= R1 ; ++ j ) {
            update ( c1[j] , c2[j] ) ;
        }
        for ( int j = L1 ; j <= R1 ; ++ j ) {
            dp[cur][j] = ( dp[cur][j - 1] + c1[j] ) % mod ;
        }
    }
    printf ( "%d\n" , dp[cur]
 ) ;
}

int main () {
    int T ;
    scanf ( "%d" , &T ) ;
    for ( int i = 1 ; i <= T ; ++ i ) {
        solve () ;
    }
    return 0 ;
}
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