HDU 5291(Candy Distribution-差值dp)

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 10⁹+7 ) in a single line.

 

Sample Input

2
1
2
2
1 2

 

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，非常easy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;
}