[hdu5389 Zero Escape]数根的性质，DP

题意：把n个数(1-9)放到A集合和B集合里面去，使得A集合里面的数的数根为a，B集合里面的数的数根为b，也可以只放在A或B任一个集合里面。求方法总数。比如A={2,4,5}，则A的数根为[2+4+5]=[11]=[2]=2

思路：一个数为a，则它的数根b=(a-1)%9+1=(digit-1)%9+1，digit是a的十进制各位上的数的和。如果存在解，那么任选一些数放到A集合里面，使得A集合的数根为a，那么B集合的数根一定为b。由公式可知，数根可以转化为余数来做，令dp[i][x]表示考虑前i个数，使得A里面数的和对9的余数为x的方法总数，则有dp[i][x]=dp[i-1][x]+dp[i-1][(x-a[i]+9)%9]。最后需要考虑只放一个集合的情况。

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define X                   first
#define Y                   second
#define pb                  push_back
#define mp                  make_pair
#define all(a)              (a).begin(), (a).end()
#define fillchar(a, x)      memset(a, x, sizeof(a))
#define copy(a, b)          memcpy(a, b, sizeof(a))

typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long ull;

//#ifndef ONLINE_JUDGE
void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);}
void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>
void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1;
while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>
void print(const T t){cout<<t<<endl;}template<typename F,typename...R>
void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>
void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}
//#endif
template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}
template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}
template<typename T>
void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];}
template<typename T>
void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];}

const double PI = acos(-1.0);
const int INF = 1e9 + 7;
const double EPS = 1e-8;

/* -------------------------------------------------------------------------------- */
template<int mod>
struct ModInt {
    const static int MD = mod;
    int x;
    ModInt(ll x = 0): x(x % MD) {}
    int get() { return x; }

    ModInt operator + (const ModInt &that) const { int x0 = x + that.x; return ModInt(x0 < MD? x0 : x0 - MD); }
    ModInt operator - (const ModInt &that) const { int x0 = x - that.x; return ModInt(x0 < MD? x0 + MD : x0); }
    ModInt operator * (const ModInt &that) const { return ModInt((long long)x * that.x % MD); }
    ModInt operator / (const ModInt &that) const { return *this * that.inverse(); }

    ModInt operator += (const ModInt &that) { x += that.x; if (x >= MD) x -= MD; }
    ModInt operator -= (const ModInt &that) { x -= that.x; if (x < 0) x += MD; }
    ModInt operator *= (const ModInt &that) { x = (long long)x * that.x % MD; }
    ModInt operator /= (const ModInt &that) { *this = *this / that; }

    ModInt inverse() const {
        int a = x, b = MD, u = 1, v = 0;
        while(b) {
            int t = a / b;
            a -= t * b; std::swap(a, b);
            u -= t * v; std::swap(u, v);
        }
        if(u < 0) u += MD;
        return u;
    }

};
typedef ModInt<258280327> mint;

const int maxn = 1e5 + 7;

mint dp[maxn][10];

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
    int T, n, a, b, x;
    cin >> T;
    while (T --) {
        cin >> n >> a >> b;
        dp[0][0] = 1;
        a = a % 9;
        b = b % 9;
        int tot = 0;
        for (int i = 1; i <= n; i ++) {
            scanf("%d", &x);
            for (int j = 0; j < 9; j ++) {
                dp[i][j] = dp[i - 1][j] + dp[i - 1][(j - x + 9) % 9];
            }
            tot = (tot * 10 + x) % 9;
        }
        mint ans = 0;
        if (tot == a && b > 0) ans += 1;
        if (tot == b && a > 0) ans += 1;
        if(tot == (a + b) % 9) ans += dp[n][a].get();
        printf("%d\n", ans.get());
    }
    return 0;
}