传送门

Zero Escape

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1460    Accepted Submission(s): 716

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

Author

SXYZ

Source

2015 Multi-University Training Contest 8

题解：

先转一发题解：http://blog.csdn.net/david_jett/article/details/47617099

解题：

比赛的时候是水过的，不得不吐槽下多校的数据真水，听说样例都没过也能A。之前发了错误的题解，实在是抱歉，现在改正了。之前大致思路是对的:

dp[i][j]=dp[i-1][j]+dp[i-1][tmp](tmp为j-arr[i],若小于等于0，需加9调整)

dp[i][j]含义为取前i个数字中的若干个，按给定规则，算出来的和为j的方案数

之前有两个细节没考虑到，

一是、当前位和我dp[i][j]中的j值相同，那么对应三种情况。

1.当前位不取，直接取前面dp[i-1][j]的值

2.当前位取，前面取的是dp[i-1][9]的值（因为加9，不变）

3.当前位取，前面都不取，此种情况特殊，直接加一即可。

其实2、3两种情况是一个数分别对应dp[i-1][0]和dp[i-1][9]的特殊情况。

二是、当a+b的和和sum相同时，最后输出结果的时候，按理说只要输出dp[n][a]就可以了，但是这样会遗漏a中都不取，全都放在b中，而b又刚好和sum相同的情况。

现在是没有错误了，请放心看。

注意点:

开始想到dp了，不过傻傻的用了3维dp，结果超时了，没注意到一部分人进了A门，那么剩下的人肯定要进B门，所以第三维可以省掉。

转移方程：

    for(o = 1;o <= n;o++){
        for(i = 0;i <= 9;i++){
            t = (9+i-x[o]) % 9;
            dp[o][i] = (dp[o-1][t] + dp[o-1][i]) %mod;
        }
    }

代码：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <algorithm>
 5 #include <stack>
 6 #include <cctype>
 7 #include <vector>
 8 #include <cmath>
 9 #include <map>
10 #include <queue>
11
12 #define ll long long
13 #define N 100005
14 #define mod 258280327
15
16 using namespace std;
17
18 int T;
19 int n,a,b;
20 int x[N];
21 int dp[N][11];
22 int ans;
23 int s;
24
25 int root(int x) {return (x-1)%9+1;}
26
27 void solve()
28 {
29     int i,o;
30     int t;
31     dp[0][0] = 1;
32     for(o = 1;o <= n;o++){
33         for(i = 0;i <= 9;i++){
34             t = (9+i-x[o]) % 9;
35             dp[o][i] = (dp[o-1][t] + dp[o-1][i]) %mod;
36         }
37     }
38 }
39
40 int main()
41 {
42     //freopen("in.txt","r",stdin);
43     scanf("%d",&T);
44     for(int ccnt=1;ccnt<=T;ccnt++){
45     //while(scanf("%d%s%s",&n,a,b)!=EOF){
46         scanf("%d%d%d",&n,&a,&b);
47         memset(dp,0,sizeof(dp));
48         int i;
49         s = 0;
50         for(i = 1;i <= n;i++){
51             scanf("%d",&x[i]);
52             s = root(s+x[i]);
53         }
54         solve();
55         ans = 0;
56         if(root(a+b)==s) ans+=dp[n][a];
57         else if(a==s) ++ans;
58         if(b==s) ++ans;
59         printf("%d\n",ans%mod);
60     }
61     return 0;
62 }