BZOJ 4800: [Ceoi2015]Ice Hockey World Championship meet_in_the_middle

4800: [Ceoi2015]Ice Hockey World Championship

Time Limit: 10 Sec  Memory Limit: 256 MB
Submit: 439  Solved: 237
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Description

有n个物品，m块钱，给定每个物品的价格，求买物品的方案数。

Input

第一行两个数n，m代表物品数量及钱数

第二行n个数，代表每个物品的价格

n<=40，m<=10^18

Output

一行一个数表示购买的方案数

（想怎么买就怎么买，当然不买也算一种）

Sample Input

5 1000
100 1500 500 500 1000

Sample Output

8

吧n的数据范围看成m的了。。。心中跑过无数草泥马

在看数据范围前想背包

看完之后又不小心GG

偷偷翻题解 meet in the middle

复杂度

#include<cmath>#include<ctime>#include<cstdio>#include<cstring>#include<cstdlib>#include<complex>#include<iostream>#include<algorithm>#include<iomanip>#include<vector>#include<string>#include<bitset>#include<queue>#include<set>#include<map>using namespace std;typedef long long ll;inline ll read(){ll x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch<='9'&&ch>='0'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}return x*f;}void print(ll x){if(x<0)putchar('-'),x=-x;if(x>=10)print(x/10);putchar(x%10+'0');}const int N=45,M=1200000;int n,midst,cnt1,cnt2;ll m,a[N],res1[M],res2[M];void dfs(int step,int aim,ll sum){if(step>aim){aim^midst?res2[++cnt2]=sum:res1[++cnt1]=sum;return ;}dfs(step+1,aim,sum);dfs(step+1,aim,sum+a[step]);}inline bool cmp(ll x,ll y){return x>y;}void solve(){ll ans=0;sort(res1+1,res1+cnt1+1);sort(res2+1,res2+cnt2+1,cmp);register int i=1,j=1;while(i<=cnt1){while(res1[i]+res2[j]>m)j++;if(j>cnt2)break;ans+=cnt2-j+1;i++;}print(ans);puts("");}int main(){n=read();m=read();for(int i=1;i<=n;++i)a[i]=read();midst=(n>>1);dfs(1,midst,0);dfs(midst+1,n,0);solve();return 0;}/*5 1000100 1500 500 500 10008*/