2015暑假多校联合---Zero Escape（变化的01背包）

题目链接

http://acm.hust.edu.cn/vjudge/contest/130883#problem/C

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

 

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wange2014

题意：输入n，A，B   表示有n个数，一部分放入A中，剩余部分放入B中，或者全放入A中、B中，A中数得满足 和的每一位相加 再对和的每一位求和......直至最后变成一位数，而这个数必须和A相等，B同样如此，求有多少种分配方案？

思路：上述对A中和B中数和的运算  等价于 和对9取余，令A中的和为suma，B中和为sumb，suma+sumb==sum  则suma%9==A   sumb%9==B  所以 如果（suma+sumb）%9==（A+B）%9  且  suma%9==A   那么必有sumb%9==B   所以问题得到简化，先判断（suma+sumb）%9==（A+B）%9  是否成立，若成立，才有可能把n个数分成两部分满足上述条件。    那么在满足的条件下，我们只需算取数放入A中满足条件的方案数，可以用01背包实现；

代码如下：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int mod=;
int a[];
int dp[][];

int main()
{
    int T,n,A,B;
    cin>>T;
    while(T--)
    {
        int sum=;
        scanf("%d%d%d",&n,&A,&B);
        memset(dp,,sizeof(dp));
        for(int i=;i<=n;i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        sum%=;
        int res=;
        if(sum==B%) res++;
        if(sum==(A+B)%)
        {
            dp[][]=;
            for(int i=;i<=n;i++)
            {
                for(int j=;j<=;j++)
                {
                    if(j>=a[i]) dp[i][j]=dp[i-][j]+dp[i-][j-a[i]];
                    else        dp[i][j]=dp[i-][j]+dp[i-][j+-a[i]];
                    dp[i][j]%=mod;
                }
            }
            printf("%d\n",(dp[n][A]+res)%mod);
        }
        else  printf("%d\n",(sum==A%)+res);
    }
    return ;
}