鏈接：https://ac.nowcoder.com/acm/contest/889/D?&headNav=acm
來源：牛客網
時間限制：C/C++ 2秒，其他語言4秒
空間限制：C/C++ 262144K，其他語言524288K
64bit IO Format: %lld
題目描述
Amy asks Mr. B problem D. Please help Mr. B to solve the following problem.
Amy wants to crack Merkle–Hellman knapsack cryptosystem. Please help it.
Given an array {ai} with length n, and the sum s.
Please find a subset of {ai}, such that the sum of the subset is s.
For more details about Merkle–Hellman knapsack cryptosystem Please read
https://en.wikipedia.org/wiki/Merkle%E2%80%93Hellman_knapsack_cryptosystem
https://blog.nowcoder.net/n/66ec16042de7421ea87619a72683f807
Because of some reason, you might not be able to open Wikipedia.
Whether you read it or not, this problem is solvable.
輸入描述:
The first line contains two integers, which are n(1 <= n <= 36) and s(0 <= s < 9 * 1018)
The second line contains n integers, which are {ai}(0 < ai < 2 * 1017).
{ai} is generated like in the Merkle–Hellman knapsack cryptosystem, so there exists a solution and the solution is unique.
Also, according to the algorithm, for any subset sum s, if there exists a solution, then the solution is unique.
輸出描述:
Output a 01 sequence. If the i-th digit is 1, then ai is in the subset. If the i-th digit is 0, then ai is not in the subset.
示例1
輸入
8 1129 295 592 301 14 28 353 120 236
輸出
01100001
說明
This is the example in Wikipedia.
示例2
輸入
36 68719476735 1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072 262144 524288 1048576 2097152 4194304 8388608 16777216 33554432 67108864 134217728 268435456 536870912 1073741824 2147483648 4294967296 8589934592 17179869184 34359738368
輸出
111111111111111111111111111111111111
題意：給你n個數和一個揹包，讓你用這n個數恰好放滿這個揹包。
思路：對於揹包問題，經典的解法是DP，複雜度爲O(nV)，顯然不能解決這個問題。而搜索解決01揹包的複雜度爲O(N*2^N)，顯然也不能很好解決這個問題。我們可以將這些物品均分爲兩份，第一份搜出所有情況之後用一個結構體保存起來，第二份搜到邊界時二分從第一份種找最符合的。這樣時間複雜度爲O(N*2^(N/2))。
#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long ll;
const int maxn=1e6+100;
int n,cnt;
ll s,a[100];
struct node{
string s;
ll val;
}ans[maxn];
void dfs1(int pos,ll val,string ss)
{
if(pos>=n/2)
{
ans[++cnt].s=ss;
ans[cnt].val=val;
return ;
}
dfs1(pos+1,val,ss+"0");
dfs1(pos+1,val+a[pos],ss+"1");
}
int flag=0;
int judge(ll val)
{
int l=1,r=cnt,flg=0;
while(l<=r)
{
int mid=(l+r)>>1;
if(ans[mid].val+val==s)
{
flg=mid;
break;
}
else if(ans[mid].val+val<s) l=mid+1;
else r=mid-1;
}
return flg;
}
void dfs2(int pos,ll val,string ss)
{
if(pos>n)
{
int pos=judge(val);
if(pos!=0)
{
flag=1;
cout<<ans[pos].s<<ss<<endl;
}
return ;
}
if(flag) return ;
dfs2(pos+1,val,ss+"0");
dfs2(pos+1,val+a[pos],ss+"1");
}
bool cmp(node x,node y)
{
return x.val<y.val;
}
int main()
{
string ss;
scanf("%d%lld",&n,&s);
for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
ss="";cnt=0;flag=0;
dfs1(1,0,ss);
ss="";
sort(ans+1,ans+1+cnt,cmp);
dfs2(n/2,0,ss);
return 0;
}