HDU 4876 ZCC loves cards _(:зゝ∠)_ 随机输出保平安

GG，，，g艹

#include <cstdio>#include <iostream>#include <algorithm>#include <string.h>#include <vector>#include <queue>#include <math.h>using namespace std;vector<int>G[21][7];//G[i][j] 表示n=i k=j的情况下 二进制的状态int n, k, l;int a[21], d[6], val[6];int vis[150], tim;int work(int x){    int i = 0, j = 0;    while(x) {        if(x&1)    d[i++] = j;//a[j]        j++;        x>>=1;    }    int ans = 0;    do    {        memset(val, 0, sizeof val);        tim ++;        int st = 0;        for(int num = 0; num < k; num++) {            i = st; j = 0;            while(1) {                val[j] ^= a[d[i]];                vis[val[j]] = tim;                i++; j++;if(i>=k)i=0;                if(i==st)break;            }            st++;        }        for(i = l; ; i++)            if(vis[i]!=tim)            {                ans = max(ans, (i-1)>=l? (i-1):0);                break;            }    } while (next_permutation(d + 1, d + k));    return ans;}void dfs(int dep, int cnt, int num) {    if (dep > 20 || cnt > 6) return ;    G[dep][cnt].push_back(num);    dfs(dep + 1, cnt + 1, num | (1 << dep));    dfs(dep + 1, cnt, num);}struct Node {    int cnt, idx;    bool operator < (const Node &rhs) const {        return cnt > rhs.cnt;    }};Node qq[24];int b[24];int cmp(int a, int b) {    return a > b;}int main(){    int i, j;    for(i = 1; i <= 20; i++)        for(j = 1; j <= 6; j++) G[i][j].clear();    dfs(0, 0, 0);    tim = 100;    while(~scanf("%d %d %d",&n, &k, &l)) {        for(i = 0; i < n; i++)scanf("%d",&a[i]);                int ans = 0, siz = (int)G[n][k].size();        if (siz <= 12000) {            for (int i = 0; i < siz; ++i) {                int cur = G[n][k][i];                ans = max(ans, work(cur));              }        } else if (siz <= 24001) {            for (int i = 1; i < siz; i += 2) {                int cur = G[n][k][i];                ans = max(ans, work(cur));            }        } else {            for (int i = 0; i < siz; i += 3) {                int cur = G[n][k][i];                ans = max(ans, work(cur));            }        }                printf("%d\n",ans);    }    return 0;}