标签:style http color os io for re c
题目大意:给出n,k,l,表示有n张牌,每张牌有值。选取其中k张排列成圈,然后在该圈上进行游戏,每次选取m(1≤m≤k)张连续的牌,取牌上值的亦或和。要求找到一个圈,使得L~R之间的数都可以得到,输出R。如果R < L输出0.
解题思路:暴力,首先预处理出来每种选取的亦或值,然后在该基础上从可以组成L的状态中挑选一个,L+1的状态中挑取一个,知道说总的挑取出所有状态中选中的牌的个数大于K为值,然后用全排序去查找最大的R。
#include <cstdio>
#include <cstring>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn = 200;
const int maxc = 25;
const int maxs = 1<<20;
const int maxt = 20005;
bool flag;
int ans, rec, dp[maxs+5];
int N, K, L, arr[maxc];
int vec[maxn][maxt], c[maxn];
inline int bitcount (int s) {
return s == 0 ? 0 : bitcount(s/2) + (s&1);
}
void dfs (int d, int s, int val, int f) {
if (d == K + 1)
return;
vec[val][c[val]++] = s;
for (int i = f + 1; i < N; i++)
dfs (d+1, s|(1<<i), val^arr[i], i);
}
void init () {
flag = true;
ans = rec = 0;
memset(c, 0, sizeof(c));
memset(dp, 0, sizeof(dp));
for (int i = 0; i < N; i++)
scanf("%d", &arr[i]);
dfs(0, 0, 0, -1);
}
int check (int* a) {
int v[maxn];
memset(v, 0, sizeof(v));
for (int m = 1; m <= K; m++) {
int tmp = a[0];
for (int i = 1; i < m; i++)
tmp ^= a[i];
for (int i = 0; i < K; i++) {
v[tmp] = 1;
tmp ^= (a[i] ^ a[i+m]);
}
}
int ans = L;
while (v[ans] && ans < maxn)
ans++;
return ans - 1;
}
void judge (int s) {
int cnt = 0, a[2*maxn];
for (int i = 0; i < N; i++) {
if (s&(1<<i))
a[cnt++] = arr[i];
}
sort(a + 1, a + cnt);
do {
for (int i = 0; i < cnt; i++)
a[i+cnt] = a[i];
ans = max(ans, check(a));
} while (next_permutation(a + 1, a + cnt));
}
/*
*/
void solve (int d, int s) {
int cnt = bitcount(s);
rec = max(d, rec);
if (cnt >= K || d >= maxn) {
if (cnt == K) {
judge(s);
flag = false;
}
return;
}
for (int i = 0; i < c[d]; i++) {
int s0 = (s | vec[d][i]);
if (dp[s0]) continue;
solve(d+1, s0);
dp[s0] = 1;
}
}
int main () {
while (scanf("%d%d%d", &N, &K, &L) == 3) {
init();
solve(L, 0);
if (flag)
ans = rec - 1;
printf("%d\n", ans < L ? 0 : ans);
}
return 0;
}
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标签:style http color os io for re c
原文地址:http://blog.csdn.net/keshuai19940722/article/details/38092257