codeforces 525 E Anya and Cubes 中途相遇法

标签：algorithm   搜索   

#include<iostream>
#include<cstdio>
#include<cstring>
#include <tr1/unordered_map>
using namespace std;
using namespace std::tr1;

#define LL unsigned __int64
#define MP make_pair
const int N=30;
const LL BEI=(LL)1e16+1;
int a[N];
unordered_map<LL,LL> mp;
LL ans=0;

int n,k;
LL s;
LL f[50];
LL fac(int x){
	if(x<50) return f[x];
	return 0;
}
void predo(int x){
	LL sum=0;
	int p=0;
	int cnt=0;
	int cx=x;
	while(x){
		int ch=x%3;
		if(ch==1) sum+=(LL)a[p];
		else if(ch==2){
			if(fac(a[p])==0) return ;
			else sum+=fac(a[p]);
			++cnt;
			if(cnt>k) return ;
		}
		if(sum>s) return ;
		x/=3;
		++p;
	}
	//cout<<"q"<<' '<<cx<<' '<<sum<<' '<<cnt<<endl;
	//++mp[MP(sum,cnt)];
	//cout<<BEI*(cnt+1)+sum<<endl;
	++mp[BEI*(cnt+1)+sum];
	if(sum==s) ++ans;
}
void gao(int n1,int x){
	LL sum=0;
	int p=n1;
	int cnt=0;
	int cx=x;
	while(x){
		int ch=x%3;
		if(ch==1) sum+=(LL)a[p];
		else if(ch==2){
			if(fac(a[p])==0) return ;
			else sum+=fac(a[p]);
			++cnt;
			if(cnt>k) return ;
		}
		if(sum>s) return ;
		x/=3;
		++p;
	}
	//cout<<cx<<' '<<sum<<' '<<cnt<<endl;
	if(s==sum) ++ans;
	else{
		for(int i=0;i<=k-cnt;++i)
			ans+=mp[BEI*(i+1)+(s-sum)];
	}
}
int main(){
	scanf("%d%d%I64u",&n,&k,&s);
	memset(f,0,sizeof(f));
	f[0]=1;
	for(int i=1;i<50;++i){
		f[i]=f[i-1]*i;
		//cout<<f[i]<<endl;
		if(f[i]>s){
			f[i]=0;
			break;
		}
	}
	for(int i=0;i<n;++i)
		scanf("%d",&a[i]);
	int n1=n/2;
	int lim1=1;
	for(int i=0;i<n1;++i) lim1*=3;
	for(int i=0;i<lim1;++i){
		predo(i);
	}
	int n2=n-n1;
	int lim2=1;
	for(int i=0;i<n2;++i) lim2*=3;
	for(int i=1;i<lim2;++i){
		gao(n1,i);
	}
	printf("%I64u\n",ans);
	return 0;
}

#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<tr1/unordered_map>
using namespace std;
using namespace std::tr1;
#define PB push_back
#define LL __int64
LL S;
LL fac[50];
int a[30];
int n,k,lim;
LL ans=0;
unordered_map<LL,int> mp;
vector<LL> res1[30];
vector<LL> res2[30];
void gao(int cur,LL sum,int e){
	if(sum>S) return ;
	if(e>k) return ;
	if(cur==lim)
		res1[e].PB(sum);
	else{
		gao(cur+1,sum,e);
		gao(cur+1,sum+a[cur],e);
		if(a[cur]<19)
			gao(cur+1,sum+fac[a[cur]],e+1);
	}
}
int main(){
	fac[0]=1;
	for(int i=1;i<19;++i){
		fac[i]=fac[i-1]*i;
	}
	scanf("%d%d%I64d",&n,&k,&S);
	for(int i=0;i<n;++i)
		scanf("%d",&a[i]);
	lim=n/2;
	gao(0,0,0);
	//for(int i=0;i<30;++i){
	//	cout<<i<<':';
	//	for(int j=0;j<res1[i].size();++j)
	//		cout<<res1[i][j]<<' ';
	//	cout<<endl;
	//}
	reverse(a,a+n);
	lim=n-lim;
	swap(res1,res2);
	gao(0,0,0);
	for(int i=k;i>=0;--i){
		for(int j=0;j<res2[k-i].size();++j){
			++mp[S-res2[k-i][j]];
		}
		for(int j=0;j<res1[i].size();++j){
			ans+=mp[res1[i][j]];
		}
	}
	printf("%I64d\n",ans);
	return 0;
}

codeforces 525 E Anya and Cubes 中途相遇法

标签：algorithm   搜索   

原文地址：http://blog.csdn.net/whai362/article/details/44829669