[AtCoder] C - Tak and Cards

传送门：https://atcoder.jp/contests/abc044/tasks/arc060_a

挺容易想到的dp，但是因为脑残改了一晚上，最后出错的地方自己都看笑了

 //数据很小，复杂度最高但最好理解的三维dp也能过
 1 #include<cstdio>
 2 #include<cstring>
 3 #include<cstdlib>
 4 #include<iostream>
 5 #include<algorithm>
 6 #include<cmath>
 7 #include<string>
 8 #include<stack>
 9 #include<queue>
10 #include<vector>
11 #include<map>
12 //#include<windows.h>
13 #define fi first
14 #define se second
15 #define pb(i) push_back(i)
16 #define rep(i,a,b) for(int i=a;i<=b;i++)
17 #define per(i,a,b) for(int i=a;i>=b;i--)
18 #define mes(a,b) memset(a,b,sizeof(a))
19 using namespace std;
20 typedef long long ll;
21 typedef unsigned long long ull;
22 const int N = 55+10;
23 const int INF = 0x3f3f3f3f;
24 const int inf = - INF;
25 const int mod = 1e9+7;
26 const double pi = acos(-1.0);
27 
28 ll x[N];
29 ll dp[N][N][N*N];
30 int main(){
31     int n,a;
32     cin>>n>>a;
33     mes(dp,0);
34     rep(i,1,n)
35         cin>>x[i];
36     rep(i,0,n){
37         rep(j,0,n){
38             rep(k,0,n*a){//i->目前到第几张牌 j->取了几张牌 k->总和
39                 if(i==0&&j==0&&k==0) dp[i][j][k]=1;
40                 else if(i>=1&&k<x[i]) dp[i][j][k]=dp[i-1][j][k];
41                 else if(i>=1&&j>=1&&k>=x[i]) dp[i][j][k]=dp[i-1][j][k]+dp[i-1][j-1][k-x[i]];
42                 else dp[i][j][k]=0;
43             }
44         }
45     }
46     ll ans=0;
47     rep(i,1,n){//最后竟然出错在这里，让i从0开始遍历了。
48         ans+=dp[n][i][i*a];
49     }
50     cout<<ans<<endl;
51     //system("pause");
52     return 0;
53 }