A. Musical Puzzle
思路:
用最少的长度为2的字符串按一定规则拼出s。规则是：前一个字符串的尾与后一个字符串的首相同。统计s中长度为2的不同字符串数量。
代码:
#include<bits/stdc++.h>
#include <unordered_map>
using namespace std;
#define N 200010
typedef long long ll;
typedef unsigned long long ull;
ll n, m, h, k, t, x, y, z;
ll a, b, c, d, mod = 998244353;
ll ans, num, sum1 = 0, sum, sum2 = 0, maxx, minn = 1e9;
ll f1[N], f2[N], dp[N], an[N], cnt[N];
bool flag, vis[N];
string s, s1, s2;
map<string, ll>mp;
bool cmp(ll x, ll y) {
return x > y;
}
void solve() {
cin >> n;
cin >> s;
ans = 0;
mp.clear();
for (int i = 0; i < s.size()-1; i++) {
s1 = s.substr(i,2);
if (!mp[s1]) {
mp[s1] = 1;
ans++;
}
}
cout << ans << endl;
}
int main()
{
cin >> t;
while (t--) {
solve();
}
return 0;
}B. Restore the Weather
思路:
将a和b分别按从小到大的顺序匹配便是最优的，一定能满足|ai−bi|≤k成立。只需注意要恢复原来的顺序输出。
代码:
#include<bits/stdc++.h>
#include <unordered_map>
using namespace std;
#define N 200010
typedef long long ll;
typedef unsigned long long ull;
//ll n, m, h, k, t, x, y, z;
//ll a, b, c, d, mod = 998244353;
//ll ans, num, cnt,sum, maxx, minn = 1e9;
//ll f1[N], f2[N], dp[N], an[N];
//bool flag, vis[N];
//string s, s1, s2;
//map<ll, ll>mp;
struct tem {
ll d, t;
}a[114514];
ll b[114514];
bool cmp1(tem a, tem b) { return a.t < b.t; }
bool cmp2(tem a, tem b) { return a.d < b.d; }
signed main()
{
ll t;
cin >> t;
while (t--) {
ll n, k;
cin >> n >> k;
for (int i = 1; i <= n; i++) {
cin >> a[i].t;
a[i].d = i;
}for (int i = 1; i <= n; i++)
cin >> b[i];
sort(a + 1, a + 1 + n, cmp1);
sort(b + 1, b + 1 + n);
for (int i = 1; i <= n; i++)a[i].t = b[i];
sort(a + 1, a + 1 + n, cmp2);
for (int i = 1; i <= n; i++)
cout << (a[i].t) << " ";
cout << "\n";
}
return 0;
}C. Vlad Building Beautiful Array
思路:
2只有减奇数才能改变奇偶性。值得注意的是，根据题目要求我们无法将一个奇数序列变成偶数序列，因此只有以下三种情况：
①原本全是奇数
②原本全是偶数
③有奇有偶：这时只能将偶序列变成奇序列。我们将每一个偶数与最小的奇数消减，只要最后的结果全为正数则有解，否则无解。
代码:
#include<bits/stdc++.h>
#include <unordered_map>
using namespace std;
#define N 200010
typedef long long ll;
typedef unsigned long long ull;
ll n, m, h, k, t, x, y, z;
ll a, b, c, d, mod = 998244353;
ll ans, num, sum1 = 0, sum, sum2 = 0, maxx, minn = 1e9;
ll f1[N], f2[N], dp[N], an[N], cnt[N];
bool flag, vis[N];
string s, s1, s2;
map<ll, ll>mp;
void solve() {
cin >> n;
minn = 1e9;
for (int i = 1; i <= n; i++) {
cin >> dp[i];
minn = min(minn, dp[i]);
}
if (minn % 2 == 0) {
for (int i = 1; i <= n; i++) {
if (dp[i] % 2 != 0) {
cout << "NO" << endl;
return;
}
}
cout << "YES" << endl;
return;
}
else {
cout << "YES" << endl;
return;
}
}
int main()
{
ios_base::sync_with_stdio(false);
cin >> t;
while (t--) {
solve();
}
return 0;
}D. Flipper
思路:
在解空间中，以n/n-1开头的排列字典序比其他可行解大，因此最优解在以n/n-1开头的解空间里。
所以本题右端点是固定的，若元素n不在开头，则r在元素n所在的位置，若n在开头，则r在元素n - 1所在的位置。固定了r后我们再去枚举左端点l，在所有可行解中取字典序最大的那个排列即可。
代码:
#include<bits/stdc++.h>
#include <unordered_map>
using namespace std;
#define N 200010
typedef long long ll;
typedef unsigned long long ull;
ll n, m, h, k, t, x, y, z;
ll a, b, c, d, mod = 998244353;
ll ans, num, cnt,sum, maxx, minn = 1e9;
ll f1[N], f2[N], dp[N], an[N];
bool flag, vis[N];
string s, s1, s2;
map<ll, ll>mp;
void solve() {
cin >> n;
for (int i = 1; i <= n; i++)
cin >> dp[i];
if (n == 1) {
cout << dp[1] << endl;
return;
}
maxx = 0;
for (int i = 2; i <= n; i++) {
if (dp[i] >= maxx) {
maxx=dp[i];
cnt = i;
}
}
if (cnt == n) {
num = cnt;
for (int i = cnt - 1; i >= 2; i--) {
if (dp[i] > dp[1]) {
num = i;
}
else
break;
}
cout << dp[cnt] << " ";
for (int i = cnt - 1; i >= num; i--)
cout << dp[i] << " ";
for (int i = 1; i < num; i++)
cout << dp[i] << " ";
cout << endl;
}
else {
num = cnt - 1;
for (int i = cnt-2; i >= 2; i--) {
if (dp[i] > dp[1]) {
num = i;
}
else
break;
}
for (int i = cnt; i <= n; i++)
cout << dp[i] << " ";
for (int i = cnt - 1; i >= num; i--)
cout << dp[i] << " ";
for (int i = 1; i < num; i++)
cout << dp[i] << " ";
cout << endl;
}
}
int main()
{
ios_base::sync_with_stdio(false);
cin >> t;
while (t--) {
solve();
}
return 0;
}