HDU4876:ZCC loves cards，hdu4876zcc

HDU4876:ZCC loves cards，hdu4876zcc
Problem DescriptionZCC 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. 
InputThe 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. 
OutputFor each test case, output the maximal number R. And if L can’t be obtained, output 0. 
Sample Input

4 3 12 3 4 5

 
Sample Output

7Hint ⊕ means xor  用全排列的方法來做，這我一開始還真沒想到
#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int n,k,l,r;int vis[500],a[500],tem[500],s[500];void set(int len,int sum){    vis[sum] = 1;    if(len == k)        return ;    set(len+1,sum^tem[len]);    set(len+1,sum);}int check(){    memset(vis,0,sizeof(vis));    set(0,0);    for(int i = l; i<=r; i++)        if(!vis[i])            return 0;    return 1;}void solve(){    if(!check()) return ;    int i,j;    for(i = 0; i<k; i++)        s[i] = tem[i];    do    {        memset(vis,0,sizeof(vis));        for(i = 0; i<k; i++)        {            int ans = 0;            for(j = i; j<k+i; j++)            {                ans^=s[(j%k)];                vis[ans] = 1;            }        }        for(i = l; i<=128; i++)//a[i]最大100，所以不會超過128            if(!vis[i])            {                r = max(r,i-1);                break;            }    }    while(next_permutation(s+1,s+k));}void dfs(int now,int len){    if(len == k)    {        solve();        return ;    }    for(int i = now; i<n; i++)    {        tem[len] = a[i];        dfs(i+1,len+1);    }}int main(){    int i,j;    while(~scanf("%d%d%d",&n,&k,&l))    {        for(i = 0; i<n; i++)            scanf("%d",&a[i]);        sort(a,a+n);//先排序，方便後面進行排列        r = l-1;        dfs(0,0);        if(r<l)            printf("0\n");        else            printf("%d\n",r);    }    return 0;}

ZCC:="XSTDZC#";{這句引用周K線的收盤價，公式自己建立} ZS:ZCC*1085; ZJ:ZCC*0925;

 
哪位朋友知道：ZCC、CT FMA01_160_B40_SEIZ_10盤銑刀是哪個國家生產的？

國產，株洲鑽石的，ZCC.CT是株洲鑽石切削刀具股份有限公司的英文簡稱！