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hdu 5536 Chip Factory 字典树+bitset 铜牌题

2024-08-14 02:42:59   223人阅读

Chip Factory

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1842    Accepted Submission(s): 833


Problem Description
John is a manager of a CPU chip factory, the factory produces lots of chips everyday. To manage large amounts of products, every processor has a serial number. More specifically, the factory produces 技术分享 chips today, the 技术分享-th chip produced this day has a serial number 技术分享技术分享.

At the end of the day, he packages all the chips produced this day, and send it to wholesalers. More specially, he writes a checksum number on the package, this checksum is defined as below:
技术分享技术分享技术分享技术分享技术分享技术分享技术分享技术分享技术分享技术分享技术分享技术分享技术分享技术分享技术分享技术分享技术分享技术分享

which 技术分享技术分享技术分享技术分享技术分享 are three different integers between 技术分享 and 技术分享. And 技术分享 is symbol of bitwise XOR.

Can you help John calculate the checksum number of today?
 

 

Input
The first line of input contains an integer 技术分享 indicating the total number of test cases.

The first line of each test case is an integer 技术分享, indicating the number of chips produced today. The next line has 技术分享 integers 技术分享技术分享技术分享技术分享技术分享技术分享技术分享技术分享技术分享技术分享技术分享, separated with single space, indicating serial number of each chip.

技术分享技术分享技术分享技术分享技术分享技术分享技术分享技术分享
技术分享技术分享技术分享技术分享技术分享技术分享技术分享技术分享
技术分享技术分享技术分享技术分享技术分享技术分享技术分享技术分享
There are at most 技术分享技术分享 testcases with 技术分享技术分享技术分享技术分享技术分享
 

 

Output
For each test case, please output an integer indicating the checksum number in a line.
 

 

Sample Input
231 2 33100 200 300
 

 

Sample Output
6 400

 题意：给你n个整型数(n<=1e3)，现在可以从其中选出s1,s2,s3三个数字，要求求出(s1+s2)^s3的最大值，至多

1e3组测试数据。

#include <cstdio>#include <iostream>#include <algorithm>#include <cstring>#include <iostream>#include <cmath>#include <map>#include <bitset>#include <stack>#include <queue>#include <vector>#include <bitset>#include <set>#define MM(a,b) memset(a,b,sizeof(a));#define inf 0x3f3f3f3fusing namespace std;typedef long long ll;#define CT continue#define SC scanfconst int N=1e3+10;int ch[40*N][3];int a[N],sz,num[40*N],val[40*N];bitset<32> v;void add(int k){   int p=0;   for(int i=30;i>=0;i--){      int id=v[i];      if(ch[p][id]==-1)  ch[p][id]=++sz;      p=ch[p][id];      num[p]+=k;   }   if(k!=-1) val[p]=v.to_ullong();//返回bitset的值}int query(){   int p=0;   for(int i=30;i>=0;i--){      int id=v[i],ano=ch[p][!id];      if(ano!=-1&&num[ano]!=0) p=ano;      else p=ch[p][id];   }   return val[p];}int main(){    int cas,n;    SC("%d",&cas);    while(cas--){        MM(ch,-1);MM(num,0);sz=0;        SC("%d",&n);        for(int i=1;i<=n;i++){            SC("%d",&a[i]);            v=a[i];            add(1);        }        int ans=0;        for(int i=1;i<=n;i++){            v=a[i];add(-1);            for(int j=i+1;j<=n;j++){               v=a[j];add(-1);               v=a[i]+a[j];               ans=max(ans,(a[i]+a[j])^query());               v=a[j];add(1);            }            v=a[i];add(1);        }        printf("%d\n",ans);    }    return 0;}

　错因：刚开始想到的是枚举下s1+s2，然后，，插入s3，，，但是发现根本处理不了重合的情况，，，。。

解答:其实先依次添加元素建好字典树后，，可以n^2的枚举s1+s2，同时在字典树中抹去这两个元素，

这时再插入s3，，每次操作完成后都将s1,s2再添加进去。。

hdu 5536 Chip Factory 字典树+bitset 铜牌题


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