HDU 5291 Candy Distribution

Candy Distribution

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 544    Accepted Submission(s): 214

 

 

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

解题：动态规划+规律优化

 

1.  $定义dp[i]表示两人之间相差i个糖果的情况数$
2. 当前有a个第i种糖果，那么我们有[dp[j] = dp[j] imes (a/2 + 1) + dp[j-1] imes((a-1)/2+1)+dp[j+1] imes((a-1)/2+1)+cdots + dp[j-a] imes ((a-a)/2 + 1) + dp[j+a] imes ((a-a)/2 + 1)]
3. $可以发现算出*dp[0]之后，算*dp[1]  = *dp[0] + dp[1] + dp[3] - dp[0] - dp[-2]$
4. $此时只要把[j+1,j+1+a]的奇数位置的dp值加起来 - [j-a,j]偶数位置的dp值 + *dp[0] = *dp[1]$

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 90000;
const LL mod = 1000000007;
LL dp[maxn],sum[2][maxn];
int bound[maxn],n;
int main(){
    int kase;
    scanf("%d",&kase);
    while(kase--){
        scanf("%d",&n);
        int S = 0;
        for(int i = 1; i <= n; ++i){
            scanf("%d",bound + i);
            S += bound[i];
        }
        if(S&1) S |= 1;
        memset(dp,0,sizeof dp);
        memset(sum,0,sizeof sum);
        dp[S] = 1;
        for(int i = 1,t = (S<<1); i <= n; ++i){
            sum[0][0] = dp[0];
            sum[1][0] = 0;
            for(int j = 1; j <= t; ++j){
                sum[0][j] = sum[0][j-1];
                sum[1][j] = sum[1][j-1];
                sum[j&1][j] += dp[j];
                sum[j&1][j] %= mod;
            }
            LL ret = 0;
            for(int j = 0; j <= bound[i]; ++j){
                ret += (LL)dp[j]*(((bound[i] - j)>>1) + 1);
                ret %= mod;
            }
            for(int j = 0,p = (bound[i]&1^1); j <= t; ++j){
                dp[j] = ret;
                int x = max(0,j - bound[i] - 1);
                ret += (sum[p][j + bound[i] + 1] - sum[p][j]);
                p ^= 1;
                ret -= sum[p][j] - sum[p][x];
                ret %= mod;
            }
        }
        printf("%I64d
",(dp[S] + mod)%mod);
    }
    return 0;
}

 参考这位大大的博客

$dp[j-1] imes((a-1)/2+1)$就是表示先取第i种的一个给自己，剩下的两人均分，

但是，我们不一定要全部分，所以那个1就是表示剩下的不分了，为什么乘以$(a-1)/2$，因为两个人可以都分1，都分2，都分$(a-1)/2$，共$(a-1)/2$种