类似于背包，dp[i]表示选出一些数它们的“根”是i的方法数（包括了所有人的“根”是i的情况）。然后就是要注意所有人都走另一个门的情况。
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 using namespace std; 5 6 const int MOD = 258280327; 7 const int N = 100001; 8 const int M = 10; 9 int id[N]; 10 int dp[M]; 11 int help[M]; 12 13 int main () 14 { 15 int t; 16 scanf("%d", &t); 17 while ( t-- ) 18 { 19 int n, a, b, sum = 0; 20 scanf("%d%d%d", &n, &a, &b); 21 for ( int i = 1; i <= n; i++ ) 22 { 23 scanf("%d", id + i); 24 sum += id[i]; 25 if ( sum > 9 ) sum -= 9; 26 } 27 int ab = a + b; 28 if ( ab > 9 ) ab -= 9; 29 if ( ab == sum ) 30 { 31 memset( dp, 0, sizeof(dp) ); 32 for ( int i = 1; i <= n; i++ ) 33 { 34 for ( int j = 1; j < M; j++ ) 35 { 36 int tmp = j + id[i]; 37 if ( tmp > 9 ) tmp -= 9; 38 help[tmp] = dp[j]; 39 } 40 for ( int j = 1; j < M; j++ ) 41 { 42 int tmp = j + id[i]; 43 if ( tmp > 9 ) tmp -= 9; 44 dp[tmp] += help[tmp]; 45 if ( dp[tmp] >= MOD ) dp[tmp] -= MOD; 46 } 47 dp[id[i]]++; 48 if ( dp[id[i]] >= MOD ) dp[id[i]] -= MOD; 49 } 50 //所有人都走b门的情况 51 if ( b == sum ) 52 { 53 printf("%d ", ( dp[a] + 1 ) % MOD); 54 } 55 else 56 { 57 printf("%d ", dp[a]); 58 } 59 } 60 else 61 { 62 if ( sum == a || sum == b ) 63 { 64 if ( a == b ) printf("2 "); 65 else printf("1 "); 66 } 67 else 68 { 69 printf("0 "); 70 } 71 } 72 } 73 return 0; 74 }