hdu5269 Chip Factory

地址:http://acm.split.hdu.edu.cn/showproblem.php?pid=5536

题目:

Chip Factory

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2044    Accepted Submission(s): 919

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 n chips today, the i-th chip produced this day has a serial number si.

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:

maxi,j,k(si+sj)⊕sk

which i,j,k are three different integers between 1 and n. And ⊕ is symbol of bitwise XOR.

Can you help John calculate the checksum number of today?

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

The first line of each test case is an integer n, indicating the number of chips produced today. The next line has n integers s1,s2,..,sn, separated with single space, indicating serial number of each chip.

1≤T≤1000
3≤n≤1000
0≤si≤109
There are at most 10 testcases with n>100

 
Output
For each test case, please output an integer indicating the checksum number in a line.
 
Sample Input
2
3
1 2 3
3
100 200 300
 
Sample Output
6
400
 思路:这题居然可以N^3爆过去,我也是惊呆了==,数据太水了吧。正解是01trie。
暴力代码:
#include<stdio.h>
int a[];
int max(int ta,int b)
{
if(ta>b)
return ta;
return b;
}
int sc(int ta,int tb,int tc,int ans)
{
return max(ans,(ta+tb)^tc);
}
int main(void)
{
int t,n;
scanf("%d",&t);
while(t--)
{
int ans=;
scanf("%d",&n);
for(int i=;i<=n;i++)
scanf("%d",&a[i]);
for(int i=;i<=n;i++)
for(int j=i+;j<=n;j++)
for(int k=j+;k<=n;k++)
ans=sc(a[i],a[j],a[k],ans),ans=sc(a[j],a[k],a[i],ans),ans=sc(a[i],a[k],a[j],ans);
printf("%d\n",ans);
}
return ;
}

01trie代码:

#include <stdio.h>
#include <string.h>
struct Trie
{
int root, tot, next[][], cnt[], end[]; inline int Newnode()
{
memset(next[tot], -, sizeof(next[tot]));
cnt[tot] = ;
end[tot] = ;
return tot ++;
} inline void Init()
{
tot = ;
root = Newnode();
} inline void Insert(int x)
{
int p = root;
cnt[p] ++;
for(int i = ; i >= ; i --)
{
int idx = (( << i) & x) ? : ;
if(next[p][idx] == -)
next[p][idx] = Newnode();
p = next[p][idx];
cnt[p] ++;
}
end[p] = x;
} inline void Del(int x)
{
int p = root;
cnt[p] --;
for(int i = ; i >= ; i --)
{
int idx = (( << i) & x) ? : ;
p = next[p][idx];
cnt[p] --;
}
} inline int Search(int x)
{
int p = root;
for(int i = ; i >= ; i --)
{
int idx = (( << i) & x) ? : ;
if(idx == )
{
if(next[p][] != - && cnt[next[p][]])
p = next[p][];
else
p = next[p][];
}
else
{
if(next[p][] != - && cnt[next[p][]])
p = next[p][];
else
p = next[p][];
}
}
return x ^ end[p];
}
}tr;
int max(int ta,int tb)
{
return ta>=tb?ta:tb;
}
int a[];
int main(void)
{
int t;
scanf("%d",&t);
while(t--)
{
int n,ans=;scanf("%d",&n);
tr.Init();
for(int i=;i<=n;i++)
scanf("%d",&a[i]),tr.Insert(a[i]);
for(int i=;i<=n;i++)
for(int j=i+;j<=n;j++)
{
tr.Del(a[i]),tr.Del(a[j]);
ans=max(ans,tr.Search(a[i]+a[j]));
tr.Insert(a[i]),tr.Insert(a[j]);
}
printf("%d\n",ans);
}
return ;
}
上一篇:python 函数式编程之lambda( ), map( ), reduce( ), filter( )


下一篇:navigator.geolocation例子