题目：

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

Author

SXYZ

Source

2015 Multi-University Training Contest 8

Recommend

wange2014

题意：有n个人，每个人有自己的编号，可以相同，有两个门也有自己的编号，可以相同，一群人能够通过一个门当且仅当他们的编号之和的数字根等于门的编号，问有多少种使得这n个人通过这两个门的方案。

思路：动态规划，01背包变种，每个人要么选A门，要么选B门，dp[i][j]表示前i个人数字根为j的方案数，那么转移方程就是dp[i][j]=dp[i-1][j]+dp[i-1][(j-idx(i)+9)%9],最终答案就是dp[n][A],注意如果可以全部从B门通过，答案应该加一。

代码：

#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include<climits>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>
using namespace std;

#define PB push_back
#define MP make_pair

#define REP(i,x,n) for(int i=x;i<(n);++i)
#define FOR(i,l,h) for(int i=(l);i<=(h);++i)
#define FORD(i,h,l) for(int i=(h);i>=(l);--i)
#define SZ(X) ((int)(X).size())
#define ALL(X) (X).begin(), (X).end()
#define RI(X) scanf("%d", &(X))
#define RII(X, Y) scanf("%d%d", &(X), &(Y))
#define RIII(X, Y, Z) scanf("%d%d%d", &(X), &(Y), &(Z))
#define DRI(X) int (X); scanf("%d", &X)
#define DRII(X, Y) int X, Y; scanf("%d%d", &X, &Y)
#define DRIII(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
#define OI(X) printf("%d",X);
#define RS(X) scanf("%s", (X))
#define MS0(X) memset((X), 0, sizeof((X)))
#define MS1(X) memset((X), -1, sizeof((X)))
#define LEN(X) strlen(X)
#define F first
#define S second
#define Swap(a, b) (a ^= b, b ^= a, a ^= b)
#define Dpoint  strcut node{int x,y}
#define cmpd int cmp(const int &a,const int &b){return a>b;}

 /*#ifdef HOME
    freopen("in.txt","r",stdin);
    #endif*/
const int MOD = 1e9+7;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<double> VD;
typedef long long LL;
typedef pair<int,int> PII;
//#define HOME

int Scan()
{
	int res = 0, ch, flag = 0;

	if((ch = getchar()) == '-')				//判断正负
		flag = 1;

	else if(ch >= '0' && ch <= '9')			//得到完整的数
		res = ch - '0';
	while((ch = getchar()) >= '0' && ch <= '9' )
		res = res * 10 + ch - '0';

	return flag ? -res : res;
}
/*----------------PLEASE-----DO-----NOT-----HACK-----ME--------------------*/

#define MAXN 100000
#define mod 258280327
int id[MAXN+5];
int dp[MAXN+5][10];
int main()
{int T;
RI(T);
while(T--)
{
    int n,A,B;
    RIII(n,A,B);
    int sum=0;
    for(int i=1;i<=n;i++)
        {RI(id[i]);
        sum+=id[i];}
    sum%=9;
    MS0(dp);
    dp[0][0]=1;
    for(int i=1;i<=n;i++)
        for(int j=0;j<=9;j++)
        dp[i][j]=(dp[i-1][((j-id[i])%9+9)%9]+dp[i-1][j])%mod;
    int ans=dp[n][A];
    if(sum==B%9)
    ans=(ans+1)%mod;
    printf("%d\n",ans);

}

        return 0;
}

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