【2020年牛客暑假第六场】K题 K-Bag

传送门:https://ac.nowcoder.com/acm/contest/5671/K

题意

T组测试,每组输入$n$和$k$,表示下一行有n个包,最多有1~k种包,下一行给出n个包的序列,问符不符合全排列的顺序,输出YES或NO(阳寿题)。

思路

因为n不一定整除k,所以前一段和后一段不一定是完整的k-全排列序列,先找出前后这一段,然后检查中间一部分是否是k-全排列(注意:k-全排列序列可重复)。当n<k时,只需要判断前一段的长度是否大于后一段长度,如果小于的话,中间一段一定不满足k-全排列序列。

==重点== 怎么判断中间是否为k-全排列序列,如果是暴力判断,可能会T,这时候就需要异或和和前缀和了,我们在输入的时候做一个前缀和、异或和预处理,再从1~k做一个异或和预处理为x,前缀和预处理为s,因为两个相同的数异或为0,所以判断一段序列中是否存在相同数字,只需要判断那一段的异或和是否等于x,这样还不够,还需要判断这一段的前缀和是否等于s,只有这两个条件满足,说明这一段序列是满足k-全排列序列。这样处理会比暴力好很多。

Code

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##include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pdd;

##define INF 0x7f7f7f
##define mem(a,b) memset(a , b , sizeof(a))
##define FOR(i, x, n) for(int i = x;i <= n; i++)

// const ll mod = 1e9 + 7;
const int MAXN = 5e5 + 10;
// const double eps = 1e-6;

ll n, k;
ll f[MAXN];
ll pre1[MAXN];
ll pre2[MAXN];
bool vis[MAXN];
map<int, int> mp;

void solve() {
scanf("%lld%lld", &n, &k);
for (int i = 1; i <= n; i++) {
scanf("%lld", &f[i]);
pre1[i] = pre1[i - 1] + f[i];
pre2[i] = pre2[i - 1] ^ f[i];
}
int tot1 = 0, tot2 = 0;
mp.clear();
for (int i = 1; i <= n; i++) { // 找前一段
if (mp[f[i]] == 1)
break;
mp[f[i]]++;
tot1 = i;
}
mp.clear();
for (int i = n; i >= 1; i--) { // 找后一段
if (mp[f[i]] == 1)
break;
mp[f[i]]++;
tot2 = i;
}
if (n < k) {
if(tot1 + 1 >= tot2)
printf("YES\n");
else
printf("NO\n");
return;
}
ll s = 0; // 总前缀和
ll x = 0; // 总异或和
for(int i = 1;i <= k; i++)
{
s += i;
x ^= i;
}
int flag = 0;
for (int len = 0; len <= tot1; len++) {
ll i = len, flag2 = 1;
for (; i + k <= n; i += k) {
ll v1 = pre1[i + k] - pre1[i];
ll v2 = pre2[i + k] ^ pre2[i];
if (s != v1 x != v2) {
flag2 = 0;
break;
}
}
if (flag2 && i + 1 >= tot2) {
flag = 1;
break;
}
}
if(flag)
printf("YES\n");
else
printf("NO\n");
}

signed main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
##ifdef ACM_LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
long long test_index_for_debug = 1;
char acm_local_for_debug;
while (cin >> acm_local_for_debug) {
if (acm_local_for_debug == '$') exit(0);
cin.putback(acm_local_for_debug);
if (test_index_for_debug > 20) {
throw runtime_error("Check the stdin!!!");
}
auto start_clock_for_debug = clock();
solve();
auto end_clock_for_debug = clock();
cout << "Test " << test_index_for_debug << " successful" << endl;
cerr << "Test " << test_index_for_debug++ << " Run Time: "
<< double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;
cout << "--------------------------------------------------" << endl;
}
##else
int T;
scanf("%d",&T);
while (T--)
solve();
##endif
return 0;
}

本文作者:jujimeizuo
本文地址https://blog.jujimeizuo.cn/2020/08/01/2020-nowcoder-shujia-6-k/
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