hdu 5389 Zero Escape dp

题意：n个数分成两堆，每堆的原根之和的原根分别要等于给出的AB，原根如：65536
是7,
6+5+5+3+6=25
然后2+5=7.

将其看成背包，则mod 9之后的那个值看成容量，做次背包就行了。

//#include <bits/stdc++.h>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <time.h>
#include <vector>
#include <cstdio>
#include <string>
#include <iomanip>
///cout << fixed << setprecision(13) << (double) x << endl;
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
#define ls rt << 1
#define rs rt << 1 | 1
#define pi acos(-1.0)
#define eps 1e-8
#define Mp(a, b) make_pair(a, b)
#define asd puts("asdasdasdasdasdf");
#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef pair <int, int> pl;
//typedef __int64 LL;
const int inf = 0x3f3f3f3f;
const int N = 100050;
const ll mod = 258280327;

ll dp
[10];
int a
;

int f( int x, int y )
{
    int tmp = x+y;
    tmp %= 9;
    return tmp ? tmp: 9;
}

int main()
{
    int A, B, n;
    int tot;
    scanf("%d", &tot);
    while( tot-- ) {
        scanf("%d%d%d", &n, &A, &B);
        memset( dp, 0, sizeof( dp ) );
        int sum = 0;
        for( int i = 1; i <= n; ++i ) {
            scanf("%d", &a[i]);
            sum = f( sum, a[i] );
        }
        for( int i = 1; i <= n; ++i )
            dp[i][a[i]] = 1;
        for( int i = 1; i <= n; ++i ) {
            for( int j = 0; j <= 9; ++j ) {
                dp[i][j] = ( dp[i][j] + dp[i-1][j] ) % mod;
                int k = f( a[i], j );
                dp[i][k] = (dp[i][k] + dp[i-1][j]) % mod;
            }
        }
        ll ans = 0;
        if( f( A, B ) == sum )
            ans = dp
[A];
        if( B == sum )
            ans++;
        printf("%lld\n", ans);
    }
    return 0;
}

题目数据弱。。。