HDU 5291(Candy Distribution-差值dp)

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本文探讨了一种公平分配不同种类糖果的方法，旨在确保两个接收者获得相同数量的糖果。通过动态规划解决该问题，并提供了详细的实现代码。

Candy Distribution

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 499 Accepted Submission(s): 189

Problem Description

WY has n kind of candy, number 1-N, The i-th kind of candy has ai. WY would like to give some of the candy to his teammate Ecry and lasten. To be fair, he hopes that Ecry’s candies are as many as lasten's in the end. How many kinds of methods are there?

Input

The first line contains an integer T<=11 which is the number of test cases.
Then

T cases follow. Each case contains two lines. The first line contains one integer n(1<=n<=200). The second line contains n integers ai(1<=ai<=200)

Output

For each test case, output a single integer (the number of ways that WY can distribute candies to his teammates, modulo 10 ⁹+7 ) in a single line.

Sample Input

Sample Output

  
   2
4

   
    
     Hint
         Sample: a total of 4, (1) Ecry and lasten are not assigned to the candy; (2) Ecry and lasten each to a second kind of candy; (3) Ecry points to one of the first kind of candy, lasten points to a second type of candy; (4) Ecry points to a second type of candy, lasten points to one of the first kind of candy.

Author

FZUACM

Source

2015 Multi-University Training Contest 1

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令f[cur][j]为当前状态,表示截至第cur类糖，A比B多j个糖的方案

f[cur][j]=f[cur-1][j]*(a[i]/2)+f[cur-1][j±1]*(a[i]-1)/2+...+f[cur][j±a[i]]*1

从系数上看

a[i]=1:

f[cur-1][j]	-1	0	1
f[cur][j]	1	1	1

a[i]=2:

f[cur-1][j]	-2	-1	0	1	2
f[cur][j]	1	1	2	1	1

a[i]=3:

f[cur-1][j]	-3	-2	-1	0	1	2	3
f[cur][j]	1	1	2	2	2	1	1

我们奇偶分类讨论，很容易发现f[cur][j+1]-f[cur][j] = 某段区间奇(偶)数位区间和 - 某段区间偶(奇)数位区间和

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Forkstep(i,k,s,n) for(int i=k;i<=n;i+=s)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define MAXN (200+10)
#define MAXSA (40000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b+llabs(a+b)/F*F+F)%F;}
//ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a+b+llabs(a+b)/F*F+F)%F;}
void sub(ll &a,ll b){a=(a-b+llabs(a-b)/F*F+F)%F;}
int n;
const int M = 40500;
int a[MAXN];
ll dp[2][MAXSA*2+10000],sum[2][MAXSA*2+10000];

int main()
{
//	freopen("hdu5291.in","r",stdin);
//	freopen(".out","w",stdout);
	
	int T;cin>>T;
	while(T--) {
		MEM(dp) MEM(sum) MEM(a)
		cin>>n;
		For(i,n) scanf("%d",&a[i]);
		
		int cur=0,s=0,tot=0;
		
		dp[cur][M]=1;
		For(i,n)
		{
			if (a[i]==0) continue;
			int s=a[i];
			MEM(sum)
			Fork(k,1,M+tot+a[i]+a[i])
			{
				sum[k&1][k]=add(sum[ (k&1) ][k-1] ,dp[cur][k] );
				sum[(k&1)^1][k]=sum[(k&1)^1][k-1] ; 
			} 
						
			
			
			tot+=a[i];
			cur^=1; MEM(dp[cur])
			
			int t=M-tot;
			dp[cur][t]=1;
			
			if (s%2==0)
				Fork(k,M-tot+1,M+tot)
					{
						dp[cur][k]=dp[cur][k-1];
						upd(dp[cur][k], sum[k&1][k+s]-sum[k&1][k-1] );
						sub(dp[cur][k], sum[(k&1)^1][k-1]-sum[(k&1)^1][k-1-s-1]);
					}
			else 
				Fork(k,M-tot+1,M+tot)
					{
						dp[cur][k]=dp[cur][k-1];
						upd(dp[cur][k], sum[(k&1)^1][k+s]-sum[(k&1)^1][k-1] );
						sub(dp[cur][k], sum[k&1][k-1]-sum[k&1][k-1-s-1]);
					}
		} 
		
		
		
		printf("%lld\n",dp[cur][M]);
	}	
	
	
	return 0;
}