【HDU】5291 Candy Distribution 【DP+打标记】_hdu candy factory csdn-优快云博客

This blog presents a solution to the 'CandyDistribution' problem from HDOJ using dynamic programming. The author explores an algorithmic approach to distribute candies among participants based on certain rules and constraints, aiming for an efficient and optimal solution.

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传送门：【HDU】5291 Candy Distribution

$my~~code:$

/*************************************************************************
    > 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][N] = 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][N] ) ;
}

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