Problem Statement
Tak has N cards. On the i-th (1≤i≤N) card is written an integer xi. He is selecting one or more cards from these N cards, so that the average of the integers written on the selected cards is exactly A. In how many ways can he make his selection?
Constraints
- 1≤N≤50
- 1≤A≤50
- 1≤xi≤50
- N, A, xi are integers.
Partial Score
- 200 points will be awarded for passing the test set satisfying 1≤N≤16.
Input
The input is given from Standard Input in the following format:
N A
x1 x2 … xN
Output
Print the number of ways to select cards such that the average of the written integers is exactly A.
Sample Input 1
4 8
7 9 8 9
Sample Output 1
5
- The following are the 5 ways to select cards such that the average is 8:
- Select the 3-rd card.
- Select the 1-st and 2-nd cards.
- Select the 1-st and 4-th cards.
- Select the 1-st, 2-nd and 3-rd cards.
- Select the 1-st, 3-rd and 4-th cards.
Sample Input 2
3 8
6 6 9
Sample Output 2
0
Sample Input 3
8 5
3 6 2 8 7 6 5 9
Sample Output 3
19
Sample Input 4
33 3
3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
Sample Output 4
8589934591
- The answer may not fit into a 32-bit integer.
题意：
给了你N个数，和一个数a，
让你从这N个数中选择出一些数，并使这些数的sum和是a的倍数。
问你有多少种方案数。
思路：
我们定义dp[i][j] 代表选了i个数时，sum和为j的方案数。
初始化dp[0][0]=1
转移方程时dp[i+1][j+x]+=dp[i][j]，x为输入的数。
细节见代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=;while(b){if(b%)ans=ans*a%MOD;a=a*a%MOD;b/=;}return ans;}
inline void getInt(int* p);
const int maxn=;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
ll dp[][*];
int k;
int n;
int main()
{
//freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
//freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);
gbtb;
cin>>n>>k;
int x;
dp[][]=1ll;
repd(i,,n)
{
cin>>x;
for(int j=i-;j>=;j--)// 枚举i-1~0, 一定倒序枚举，以避免重复相加。
{
for(int w=;w<=*j;w++)// 枚举选了j个数的所有可能得到的值。
{
dp[j+][w+x]+=dp[j][w]; // 加上贡献。
}
}
}
ll ans=0ll;
repd(i,,n)
{
ans+=dp[i][i*k];// 根据平均数的公式计算答案。
}
cout<<ans<<endl; return ;
} inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '\n');
if (ch == '-') {
*p = -(getchar() - '');
while ((ch = getchar()) >= '' && ch <= '') {
*p = *p * - ch + '';
}
}
else {
*p = ch - '';
while ((ch = getchar()) >= '' && ch <= '') {
*p = *p * + ch - '';
}
}
}