Candy Distribution

题目连接：

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

题意

有n种糖果，每种糖果ai个，然后你要把糖果分给A,B两个人，问你使得两个人糖果个数都相同的方案有多少种。

题解：

最简单的dp：

dp[i][j]表示考虑到第i种糖果，现在A比B多j个，那么dp[i][x-y+j]+=dp[i-1][j]，如果x+y<=ai的话

当然这种dp肯定会tle的。

然后优化一下，发现dp[i][j]=dp[i][j-k]*((a[i]-k)/2+1)，表示你先扔了K个给A，然后剩下的水果AB平分。

这个东西我们发现，是一个分奇数和偶数的等差数列的东西。

1 1 2 2 3 3 4 4 ..... n n n n-1 n-1 ..... 3 3 2 2 1 1 系数是这样的。

显然可以分奇偶前缀和+差分+递推就可以O(1)得到每一个dp[i][j]了。

然后这道题就可以搞了。

代码

#include
    
     using namespace std;const int maxn = 8e4+5;const int mod = 1e9+7;int dp[maxn],sum[2][maxn],a[205],tot;void solve(){    memset(dp,0,sizeof(dp));    memset(sum,0,sizeof(sum));    tot=0;    int n;scanf("%d",&n);    for(int i=1;i<=n;i++)        scanf("%d",&a[i]),tot+=a[i];    if(tot%2==1)tot++;    int M=tot*2;    dp[tot]=1;    for(int i=1;i<=n;i++)    {        sum[0][0]=dp[0];        sum[1][0]=0;        for(int j=1;j<=M;j++)        {            sum[0][j]=sum[0][j-1];            sum[1][j]=sum[1][j-1];            if(j%2==1)sum[1][j]+=dp[j];            else sum[0][j]+=dp[j];            sum[0][j]%=mod;            sum[1][j]%=mod;        }        long long res = 0;        for(int j=0;j<=a[i];j++)            res = (res + 1ll*dp[j]*((a[i]-j)/2+1))%mod;        int o = (a[i]%2)^1;        for(int j=0;j<=M;j++)        {            dp[j]=res;            res+=sum[o][(j+a[i]+1)];            res%=mod;            res-=sum[o][j];            res%=mod;            o^=1;            res-=sum[o][j];            res%=mod;            res+=sum[o][max((j-a[i]-1),0)];            res%=mod;        }    }    dp[tot]=dp[tot]%mod;    dp[tot]=(dp[tot]+mod)%mod;    cout<
     
      <