Codeforces Round 960 (Div. 2) 补题记录(A~D)
1.CodeForces Round #939(Div. 2) 补题记录(A~F)2.CodeTON Round 8 (Div. 1 + Div. 2, Rated, Prizes!) 补题记录(A~A)3.Educational Codeforces Round 163 (Rated for Div. 2) 补题记录(A~D)4.CodeForces Round #951(Div. 2) 补题记录(A~E)5.Codeforces Global Round 26 补题记录(A~C2)6.CodeForces Round #959 sponsored by NEAR (Div. 1 + Div. 2) 补题记录(A~E)
7.Codeforces Round 960 (Div. 2) 补题记录(A~D)
8.Codeforces Round 961 (Div. 2) 补题记录(A~D)9.Codeforces Round 962 (Div. 3) 补题记录(A~G)10.Pinely Round 4 (Div. 1 + Div. 2) 补题记录(A~F)11.Educational Codeforces Round 168 (Rated for Div. 2) 补题记录(A~E)12.Codeforces Round 963 (Div. 2) 补题记录(A~D,F1)13.Codeforces Round 964 (Div. 4) 补题记录(A~G2)14.Codeforces Round 965 (Div. 2) 补题记录(A,B,D,E1)15.EPIC Institute of Technology Round August 2024 (Div. 1 + Div. 2) 补题记录(A~D1,E)16.Educational Codeforces Round 169 (Rated for Div. 2) 补题记录(A~F)17.Codeforces Round 972 (Div. 2) 补题记录(A~C,E1)18.Codeforces Round 988 (Div. 3) 补题记录(A~G)19.Rayan Programming Contest 2024 - Selection (Codeforces Round 989, Div. 1 + Div. 2) 补题记录(A~E)打的稀烂,但是还是上分了(
A
考虑对值域做一个后缀和。若某一个后缀和的值是奇数那么先手就可以获胜。否则就不可以获胜。
(我才不会告诉你我这题吃了一次罚时的)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define int long long
using namespace std;
int mysqrt(int x) {
int l = 0, r = x + 10, best = -1;
while (l <= r) {
int mid = l + r >> 1;
if ((__int128)(mid) * (__int128)(mid) <= x)
best = mid, l = mid + 1;
else
r = mid - 1;
}
return best;
}
const int N = 500100;
int n;
int a[N], box[N];
signed main() {
int T;
cin >> T;
while (T--) {
scanf("%lld", &n);
for (int i = 1; i <= n; ++i)
scanf("%lld", &a[i]), box[i] = 0;
for (int i = 1 ;i <= n; ++i)
++box[a[i]];
if (box[n] & 1) {
puts("YES");
continue;
}
for (int i = n - 1; i; --i) {
box[i] += box[i - 1];
if (box[i] & 1) {
puts("YES");
goto ee;
}
}
puts("NO");
ee:;
}
return 0;
}
B
构造。因为
- 对于
,考虑从后往前交替构造 ,这样可以满足比 要小的位置的后缀和不会大于 位置的后缀和。 - 对于
, 考虑全部构造 。 - 对于
,考虑正着构造 ,这样可以满足比 要大的位置的前缀和不会大于 位置的前缀和。
证明显然。时间复杂度为
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define int long long
using namespace std;
int mysqrt(int x) {
int l = 0, r = x + 10, best = -1;
while (l <= r) {
int mid = l + r >> 1;
if ((__int128)(mid) * (__int128)(mid) <= x)
best = mid, l = mid + 1;
else
r = mid - 1;
}
return best;
}
const int N = 500100;
int n;
int a[N], box[N];
signed main() {
int T;
cin >> T;
while (T--) {
int n, x, y;
scanf("%lld%lld%lld", &n, &x, &y);
int idx = 0;
for (int i = y - 1; i; --i) {
++idx;
if (idx & 1) a[i] = -1;
else a[i] = 1;
}
for (int i = y; i <= x; ++i) {
a[i] = 1;
}
idx = 0;
for (int i = x + 1; i <= n; ++i) {
++idx;
if (idx & 1) a[i] = -1;
else a[i] = 1;
}
for (int i = 1; i <= n; ++i)
printf("%lld ", a[i]);
puts("");
}
return 0;
}
C
奇怪的结论。容易发现执行
问题在于如何让
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define int long long
using namespace std;
int mysqrt(int x) {
int l = 0, r = x + 10, best = -1;
while (l <= r) {
int mid = l + r >> 1;
if ((__int128)(mid) * (__int128)(mid) <= x)
best = mid, l = mid + 1;
else
r = mid - 1;
}
return best;
}
const int N = 500100;
int n;
int a[N], box[N], b[N], s[N];
signed main() {
int T;
cin >> T;
while (T--) {
int n;
scanf("%lld", &n);
int now = 0, sum = 0;
for (int i = 1; i <= n; ++i)
scanf("%lld", &a[i]), box[i] = 0, sum += a[i];
for (int _ = 0; _ < 20; ++_) {
for (int i = 1; i <= n; ++i)
box[i] = 0;
now = 0;
for (int i = 1; i <= n; ++i) {
++box[a[i]];
if (box[a[i]] >= 2 && now < a[i])
now = a[i];
b[i] = now;
}
for (int i = 1; i <= n; ++i)
s[i] = s[i - 1] + b[i], a[i] = b[i];
// cout << _ << ": ";
// for (int i = 1; i <= n; ++i)
// cout << b[i] << ' '; cout << '\n';
sum += s[n];
}
int p = 1, rig = n;
while (p <= rig) {
set<int> se;
se.insert(b[p]);
while (p < rig && (!se.count(b[p + 1]) || b[p] == 0))
++p, se.insert(b[p]);
--rig;
// cout << "qwq " << p << ' ' << rig << ' ' << sum << '\n';
sum += s[rig] - s[p - 1];
}
printf("%lld\n", sum);
}
return 0;
}
D
分类讨论。
容易发现
容易发现如果对于一段极大的区间
然后随便 dp 一下,设
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define int long long
using namespace std;
int mysqrt(int x) {
int l = 0, r = x + 10, best = -1;
while (l <= r) {
int mid = l + r >> 1;
if ((__int128)(mid) * (__int128)(mid) <= x)
best = mid, l = mid + 1;
else
r = mid - 1;
}
return best;
}
const int N = 500100;
int n, f[N][20], g[N][20], a[N], F[N];
int askmin(int l, int r) {
int p = __lg(r - l + 1);
return min(f[l][p], f[r - (1ll << p) + 1][p]);
}
int askmax(int l, int r) {
int p = __lg(r - l + 1);
return max(g[l][p], g[r - (1ll << p) + 1][p]);
}
signed main() {
int T;
cin >> T;
while (T--) {
int n;
scanf("%lld", &n);
for (int i = 1; i <= n; ++i)
scanf("%lld", &a[i]), f[i][0] = g[i][0] = a[i];
for (int i = 1; i < 20; ++i)
for (int j = 1; j <= n - (1ll << i) + 1; ++j) {
f[j][i] = min(f[j][i - 1], f[j + (1ll << (i - 1))][i - 1]);
g[j][i] = max(g[j][i - 1], g[j + (1ll << (i - 1))][i - 1]);
}
F[0] = 0;
for (int i = 1; i <= n; i++) {
if (a[i]) {
F[i] = F[i - 1] + 1;
if (a[i] <= 2) {
int l = 1, r = i - 1, best = -1;
while (l <= r) {
int mid = l + r >> 1;
if (askmin(mid, i - 1) <= 2)
l = mid + 1, best = mid;
else
r = mid - 1;
}
if (best != -1 && askmax(best, i) <= 4 && ~(i - best + 1) & 1)
F[i] = min(F[i], F[best - 1] + (i - best + 1) - 1);
}
} else F[i] = F[i - 1];
}
printf("%lld\n", F[n]);
}
return 0;
}
本文来自博客园,作者:yhbqwq,转载请注明原文链接:https://www.cnblogs.com/yhbqwq/p/18314184,谢谢QwQ
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