HDU 5389 Zero Escape(动态规划)——多校练习8

Zero Escape

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Problem Description

Zero Escape, is a visual novel adventure video game directed by Kotaro Uchikoshi (you may hear about ever17?) and developed by Chunsoft.

Stilwell is enjoying the first chapter of this series, and in this chapter digital root is an important factor. 

This is the definition of digital root on Wikipedia:
The digital root of a non-negative integer is the single digit value obtained by an iterative process of summing digits, on each iteration using the result from the previous iteration to compute a digit sum. The process continues until a single-digit number is reached.
For example, the digital root of  65536  is  7 , because  6+5+5+3+6=25  and  2+5=7 .

In the game, every player has a special identifier. Maybe two players have the same identifier, but they are different players. If a group of players want to get into a door numbered  X(1≤X≤9) , the digital root of their identifier sum must be  X .
For example, players  {1,2,6}  can get into the door  9 , but players  {2,3,3}  can't.

There is two doors, numbered  A  and  B . Maybe  A=B , but they are two different door.
And there is  n  players, everyone must get into one of these two doors. Some players will get into the door  A , and others will get into the door  B .
For example: 
players are  {1,2,6} ,  A=9 ,  B=1
There is only one way to distribute the players: all players get into the door  9 . Because there is no player to get into the door  1 , the digital root limit of this door will be ignored.

Given the identifier of every player, please calculate how many kinds of methods are there,  mod 258280327 .

 

Input

The first line of the input contains a single number  T , the number of test cases.
For each test case, the first line contains three integers  n ,  A  and  B .
Next line contains  n  integers  idi , describing the identifier of every player.
T≤100 ,  n≤105 ,  ∑n≤106 ,  1≤A,B,idi≤9

 

Output

For each test case, output a single integer in a single line, the number of ways that these  n  players can get into these two doors.

 

Sample Input

  

   4
3 9 1
1 2 6
3 9 1
2 3 3
5 2 3
1 1 1 1 1
9 9 9
1 2 3 4 5 6 7 8 9
  

 

Sample Output

  

   1
0
10
60
  

 

Source

2015 Multi-University Training Contest 8

 

/*********************************************************************/

题意：给你n个人的id，有两扇门，每个门有一个标号，我们记作A和B，现在我们要将n个人分成两组，进入两扇门中，当且仅当进入该门的所有人的id之和的digital root(数字根)等于该门的标号方可进入，问有多少种不同的分法。

放上出题人的解题报告

在这先解释一下什么是数字根，例如65536，它的数字根为7，方法是不断将各位数字相加，直到成1位数为止，即

65536->6+5+5+3+6=25->2+5=7

因此一个数的数字根只有可能是1~9，或许你会发现一个数 模9等于该数各位数字和模9，例如 33%9=6%9；

以下是证明：

关于数字根的问题，在此推荐可以去做做HDU 1013 Digital Roots

这样之后，我们便可以直接采用动态规划(背包类 取或不取)的思想来解决问题了

令s[i][j]表示使前i个人id之和的数字根为j的方法数，那么计算s[i][j]分为两部分，一部分为前i-1个人id之和的数字根为j，不取第i个人进入该门；另一部分则是前i-1个人id之和的数字根+第i个人的id产生新的数字根的方法数

即状态转移方程为s[i][j]=(s[i-1][k]+s[i-1][j])%mod;其中k为j-a[i]，i表示第i个人，j从1到9，若算出的k小于等于0还需加上9。转移方程中的两项分别对应，当前i位取或不取。因为每个数字是由确定的两个数字组合加起来得到的，比如2，若其中一个为1，那么另外一个也一定为1。且存在这样的性质，若干个数，按给定的规则先加，或者分开加，再组合在一起，得到的结果是一样的。故可以先预先把所有的数加起来，若其和和A+B得到的结果相同，那么直接取s[n][a]+s[n][b]的值即可，若不同，那么只可以放在任意的一扇门中，分别判断一下即可。

直接上代码，若有疑问，欢迎讨论，谢谢

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<stdlib.h>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-10
using namespace std;
const int N = 100005;
const int inf = 1000000000;
const int mod = 258280327;
int a[N],s[N][10];
int digital(int x)//用于求数字根
{
    x%=9;
    if(!x)
        x=9;
    return x;
}
int main()
{
    int t,n,A,B,sum,i,j,k,ans;
    scanf("%d",&t);
    while(t--)
    {
        sum=0;
        scanf("%d%d%d",&n,&A,&B);
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            sum=digital(a[i]+sum);
        }
        if(sum==digital(A+B))//判断所有人的id之和的数字根是否与两扇门标号之和的数字根相等
        {
            memset(s,0,sizeof(s));
            s[1][a[1]]=1;
            for(i=2;i<=n;i++)
                for(j=1;j<=9;j++)
                {
                    k=j-a[i];
                    if(k<=0)
                        k+=9;
                    s[i][j]=(s[i-1][j]+s[i-1][k])%mod;
                }
            printf("%d\n",(s[n][A]+s[n][B])%mod);
        }
        else//若不相等，要么无解，要么是所有人仅能进入其中一扇门
        {
            ans=0;
            if(sum==A||sum==B)
                ans++;
            printf("%d\n",ans);
        }
    }
    return 0;
}

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