Codeforces Round #297 (Div. 2) E. Anya and Cubes

中途相遇法。。分成两半计算。。。

#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 2005
#define maxm 300005
#define eps 1e-3
#define mod 9999677
//#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid 
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
//head

map<LL, int> mpp[30];
const LL INF = 1e16;
LL Sum;
int n, K;
LL ans;
int a[105];
LL fac[30];

void dfs(int pos, LL sum, int *b, int k)
{
	if(sum > Sum) return;
	if(pos == n+1) {
		mpp[K-k][sum]++;
		return;
	}
	dfs(pos+1, sum, b, k);
	dfs(pos+1, sum+b[pos], b, k);
	if(b[pos] <= 18 && k) dfs(pos+1, sum+fac[b[pos]], b, k-1);
}

void DFS(int pos, LL sum, int *b, int k)
{
	if(sum > Sum) return;
	if(pos == n+1) {
		ans += mpp[k][Sum - sum];
		return;
	}
	DFS(pos+1, sum, b, k);
	DFS(pos+1, sum+b[pos], b, k);
	if(b[pos] <= 18 && k) DFS(pos+1, sum+fac[b[pos]], b, k-1);
}

void work()
{
	int m, k;
	scanf("%d%d%I64d", &m, &k, &Sum);
	K = k;
	for(int i = 1; i <= m; i++) scanf("%d", &a[i]);
	ans = 0;
	n = m / 2;
	dfs(1, 0, a, k);
	n = m - n;
	for(int i = 0; i < K; i++)
		for(map<LL, int>::iterator it = mpp[i].begin(); it != mpp[i].end(); it++)
			mpp[i+1][it->first] += it->second;
	DFS(1, 0, a+m/2, k);
	printf("%I64d\n", ans);
}

int main()
{
	fac[1] = 1;
	for(int i = 2; i <= 18; i++) fac[i] = fac[i-1] * i;
	work();
	
	return 0;
}