A. Yes-Yes?
#include <bits/stdc++.h>
using namespace std;
#define int long long
using pii = pair<int, int>;
using vi = vector<int>;
const string T = "Yes";
void solve() {
string s;
cin >> s;
int i = -1;
if( s[0] == 'Y' ) i = 0;
else if( s[0] == 'e' ) i = 1;
else if( s[0] == 's' ) i = 2;
if( i == -1 ){
cout << "NO\n";
return ;
}
for( auto c : s ){
if( c != T[i] ){
cout << "NO\n";
return ;
}
i = ( i + 1 ) % 3;
}
cout << "YES\n";
return;
}
int32_t main() {
ios::sync_with_stdio(false), cin.tie(nullptr);
int t;
cin >> t;
while (t--)
solve();
return 0;
}
B. Lost Permutation
枚举序列的最大值
#include <bits/stdc++.h>
using namespace std;
#define int long long
using pii = pair<int, int>;
using vi = vector<int>;
void solve() {
int m, s, b = 0;
cin >> m >> s;
for (int i = 1, x; i <= m; i++)
cin >> x, b = max(b, x), s += x;
while (b * (b + 1) / 2 < s) b++;
cout << (b * (b + 1) / 2 == s ? "YES\n" : "NO\n");
return;
}
int32_t main() {
ios::sync_with_stdio(false), cin.tie(nullptr);
int t;
cin >> t;
while (t--)
solve();
return 0;
}
C. Thermostat
情况数量不多
\[a=b\\ a\rightarrow b\\ a \rightarrow l \rightarrow b\\ a \rightarrow r \rightarrow b\\ a \rightarrow l \rightarrow r \rightarrow b\\ a \rightarrow r \rightarrow l \rightarrow b \]逐个判断就好了
#include <bits/stdc++.h>
using namespace std;
#define int long long
using pii = pair<int, int>;
using vi = vector<int>;
void solve() {
int l, r, a, b, x;
cin >> l >> r >> x >> a >> b;
bool ab = abs(a - b) >= x;
bool al = abs(a - l) >= x;
bool ar = abs(a - r) >= x;
bool lr = abs(l - r) >= x;
bool lb = abs(l - b) >= x;
bool rb = abs(r - b) >= x;
if (a == b)cout << "0\n";
else if (ab) cout << "1\n";
else if (al and lb) cout << "2\n";
else if (ar and rb) cout << "2\n";
else if (al and lr and rb) cout << "3\n";
else if (ar and lr and lb) cout << "3\n";
else cout << "-1\n";
return;
}
int32_t main() {
ios::sync_with_stdio(false), cin.tie(nullptr);
int t;
cin >> t;
while (t--)
solve();
return 0;
}
D. Make It Round
对于一个数字\(x=2^a\times5^b\times c\)，那么末尾0的数量就是\(\min(a,b)\)
所以我们枚举乘数中2 和5的次方数就好了
#include <bits/stdc++.h>
using namespace std;
#define int long long
using pii = pair<int, int>;
using vi = vector<int>;
void solve() {
int n, m, res, cnt, a = 0, b = 0;
cin >> n >> m;
for (int t = n; t > 0 and t % 2 == 0; t /= 2, a++);
for (int t = n; t > 0 and t % 5 == 0; t /= 5, b++);
res = n, cnt = min(a, b);
for (int i = 0, x = 1; x <= m; i++, x *= 2)
for (int j = 0, y = 1; x * y <= m; j++, y *= 5) {
int t = min(a + i, b + j), p = m / (x * y) * (x * y) * n;
if (t == cnt and p > res) res = p;
else if (t > cnt) cnt = t, res = p;
}
if (cnt == min(a, b)) res = n * m;
cout << res << "\n";
return;
}
int32_t main() {
ios::sync_with_stdio(false), cin.tie(nullptr);
int t;
cin >> t;
while (t--)
solve();
return 0;
}
E. The Humanoid
使用药剂的顺序只有三种，三种顺序每种都贪心的取一下，最后取最优解即可。
#include <bits/stdc++.h>
using namespace std;
#define int long long
using pii = pair<int, int>;
using vi = vector<int>;
int n, h;
vi a;
int ggb() {
int i = 0, w = h;
while (i < n and a[i] < w)
w += a[i] / 2, i++;
w *= 2;
while (i < n and a[i] < w)
w += a[i] / 2, i++;
w *= 2;
while (i < n and a[i] < w)
w += a[i] / 2, i++;
w *= 3;
while (i < n and a[i] < w)
w += a[i] / 2, i++;
return i;
}
int gbg() {
int i = 0, w = h;
while (i < n and a[i] < w)
w += a[i] / 2, i++;
w *= 2;
while (i < n and a[i] < w)
w += a[i] / 2, i++;
w *= 3;
while (i < n and a[i] < w)
w += a[i] / 2, i++;
w *= 2;
while (i < n and a[i] < w)
w += a[i] / 2, i++;
return i;
}
int bgg() {
int i = 0, w = h;
while (i < n and a[i] < w)
w += a[i] / 2, i++;
w *= 3;
while (i < n and a[i] < w)
w += a[i] / 2, i++;
w *= 2;
while (i < n and a[i] < w)
w += a[i] / 2, i++;
w *= 2;
while (i < n and a[i] < w)
w += a[i] / 2, i++;
return i ;
}
void solve() {
cin >> n >> h;
a = vi(n);
for (auto &i: a) cin >> i;
sort(a.begin(), a.end());
cout << max({ggb(), bgg(), gbg()}) << "\n";
return;
}
int32_t main() {
ios::sync_with_stdio(false), cin.tie(nullptr);
int t;
cin >> t;
while (t--)
solve();
return 0;
}
F. All Possible Digits
其实我们只能对末尾进行操作，且只能加一。
末尾数字是\(pn\)，我们只用考虑两种情况
- \([0,pn]\)中所有的数字都已经出现过了，我们只需要找到\((pn,p)\)中没出现过的最大值，然后加过去即可。
- \([0,pn]\)中所有的数字没有都出现过，首先一直加直到进位，此时\((pn,p)\)中所有的都已经出现过，当前末尾是0，找到\((0,pn)\)中没有出现过的最大值，加过去就好
#include <bits/stdc++.h>
using namespace std;
#define int long long
using pii = pair<int, int>;
using vi = vector<int>;
void solve() {
int n, p;
cin >> n >> p;
vector<int> a(n);
for (auto &i: a) cin >> i;
int pn = a.back();
auto b = a;
sort(a.begin(), a.end());
a.resize(unique(a.begin(), a.end()) - a.begin());
if (pn < a.size() and a[pn] == pn) {
p--;
while (p == a.back() and p > pn)
p--, a.pop_back();
cout << p - pn << "\n";
} else {
reverse(b.begin(), b.end());
int res = p - pn;
b[0] = p;
for (int i = 0; i < n - 1; i++) {
if (b[i] < p) continue;
b[i + 1] += b[i] / p, b[i] %= p;
}
for (int x; b.back() >= p;)
x = b.back() / p, b.back() %= p, b.push_back(x);
for (auto i: b)
a.push_back(i);
sort(a.begin(), a.end());
a.resize(unique(a.begin(), a.end()) - a.begin());
auto it = lower_bound(a.begin(), a.end(), pn);
while (*it == pn and pn >= 0)
it--, pn--;
res += max(pn, 0ll);
cout << res << "\n";
}
return;
}
int32_t main() {
ios::sync_with_stdio(false), cin.tie(nullptr);
int t;
cin >> t;
while (t--)
solve();
return 0;
}
G. Restore the Permutation
对于相邻的两个数，取最大值，所以一定是把当前所有数字放在偶数位上，然后在奇数位置放一个较小的数字。
首先我们判断时候有解。
我们把所有没有放置的数字存起来，然后正序操作，每次放置一个尽可能大的，如果不能全部放完，则无解。
然后找字典序最小的解，反序操作一遍即可。
然后其实发现，我们直接反序操作就行，放不完直接就可以判断出无解。
#include <bits/stdc++.h>
using namespace std;
#define int long long
using pii = pair<int, int>;
using vi = vector<int>;
void solve() {
int n;
cin >> n;
vi vis(n + 1), res(n + 1);
for (int i = 2; i <= n; i += 2)
cin >> res[i], vis[res[i]]++;
if (*max_element(vis.begin(), vis.end()) > 1) {
cout << "-1\n";
return;
}
set<int> a;
for (int i = 1; i <= n; i++)
if (vis[i] == 0) a.insert(i);
for (int i = n; i >= 1; i -= 2) {
auto it = a.lower_bound(res[i]);
if (it == a.begin()) {
cout << "-1\n";
return;
}
it = prev(it), res[i - 1] = *it, a.erase(it);
}
for( int i = 1 ; i <= n ; i ++ )
cout << res[i] << " ";
cout << "\n";
return;
}
int32_t main() {
ios::sync_with_stdio(false), cin.tie(nullptr);
int t;
cin >> t;
while (t--)
solve();
return 0;
}