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ZCC loves cards
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2362 Accepted Submission(s): 590
Problem Description
ZCC loves playing cards. He has n magical cards and each has a number on it. He wants to choose k cards and place them around in any order to form a circle. He can choose any several
consecutive cards the number of which is m(1<=m<=k) to play a magic. The magic is simple that ZCC can get a number x=a1⊕a2…⊕am, which ai means the number on the ith card he chooses. He can play the magic infinite times, but
once he begin to play the magic, he can’t change anything in the card circle including the order.
ZCC has a lucky number L. ZCC want to obtain the number L~R by using one card circle. And if he can get other numbers which aren’t in the range [L,R], it doesn’t matter. Help him to find the maximal R.
consecutive cards the number of which is m(1<=m<=k) to play a magic. The magic is simple that ZCC can get a number x=a1⊕a2…⊕am, which ai means the number on the ith card he chooses. He can play the magic infinite times, but
once he begin to play the magic, he can’t change anything in the card circle including the order.
ZCC has a lucky number L. ZCC want to obtain the number L~R by using one card circle. And if he can get other numbers which aren’t in the range [L,R], it doesn’t matter. Help him to find the maximal R.
Input
The input contains several test cases.The first line in each case contains three integers n, k and L(k≤n≤20,1≤k≤6,1≤L≤100). The next line contains n numbers means the numbers on the n cards. The ith number a[i] satisfies 1≤a[i]≤100.
You can assume that all the test case generated randomly.
You can assume that all the test case generated randomly.
Output
For each test case, output the maximal number R. And if L can’t be obtained, output 0.
Sample Input
4 3 1 2 3 4 5
Sample Output
7Hint⊕ means xor
Author
镇海中学
Source
题意：给n个数。从中选出k个数，这k个数能够随意排列，一旦定了顺序就不能改变，在这个确定的顺序。选出m（m<=k）个数异或得到的值x,在这个顺序得到的全部x的值中找出一个最大值R，这些数中使得存在一个连续的范围L~R。
#include<stdio.h>
#include<string.h>
int n,k,L,ans[25];
int a[13],aa[13],R,flag[150];
int vist[10];
void find(int tk)
{
if(tk==k-1)
{
memset(flag,0,sizeof(flag));
for(int i=0;i<k-1;i++)
a[i+k]=a[i];
int maxa=0;
for(int i=0;i<k;i++)//枚举一个确定序列的连续片断的异或值
{
int x=a[i]; flag[x]=1; if(maxa<x)maxa=x;
for(int j=i+1;j-i+1<=k;j++)
{
x^=a[j]; flag[x]=1;if(maxa<x)maxa=x;
}
}
int r=0;
for(int i=L;i<=maxa;i++)//找出这个最大值R,使得这些数存在L~R范围的数都存在。
if(flag[i]==0)break;
else r=i;
if(r>R)R=r;
return ;
}
tk++;
for(int i=0;i<k;i++)
if(vist[i]==0)
{
a[tk]=aa[i]; vist[i]=1; find(tk); vist[i]=0;
}
}
bool panduan()//放宽条件（随意顺序）推断
{
memset(flag,0,sizeof(flag));
int maxa=0;
for(int i=1;i<(1<<k);i++)
{
int x=0;
for(int j=0;(1<<j)<=i;j++)
if((1<<j)&i)x^=aa[j];
flag[x]=1;
if(maxa<x)maxa=x;
}
int r=0;
for(int i=L;i<=maxa;i++)
if(flag[i]==0)break;
else r=i;
return r>R;
}
void CNM(int tk,int i)
{
if(tk==k-1)
{
if(panduan())
find(-1);
return ;
}
tk++;
for(int j=i+1;j<n;j++)
{
aa[tk]=ans[j]; CNM(tk,j);
}
}
int main()
{
while(scanf("%d%d%d",&n,&k,&L)>0)
{
R=0; memset(vist,0,sizeof(vist));
for(int i=0;i<n;i++)
scanf("%d",&ans[i]);
CNM(-1,-1);
printf("%d\n",R);
}
}
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