CF1398F-Controversial Rounds
题目:
题目描述:
Alice and Bob play a game. The game consists of several sets, and each set consists of several rounds. Each round is won either by Alice or by Bob, and the set ends when one of the players has won $ x $ rounds in a row. For example, if Bob won five rounds in a row and $ x = 2 $ , then two sets ends.
You know that Alice and Bob have already played $ n $ rounds, and you know the results of some rounds. For each $ x $ from $ 1 $ to $ n $ , calculate the maximum possible number of sets that could have already finished if each set lasts until one of the players wins $ x $ rounds in a row. It is possible that the last set is still not finished — in that case, you should not count it in the answer.
输入格式:
The first line contains one integer $ n $ ( $ 1 \le n \le 10^6 $ ) — the number of rounds.
The second line contains one string $ s $ of length $ n $ — the descriptions of rounds. If the $ i $ -th element of the string is 0, then Alice won the $ i $ -th round; if it is 1, then Bob won the $ i $ -th round, and if it is ?, then you don’t know who won the $ i $ -th round.
输出格式:
In the only line print $ n $ integers. The $ i $ -th integer should be equal to the maximum possible number of sets that could have already finished if each set lasts until one of the players wins $ i $ rounds in a row.
样例:
样例输入 1:
样例输出 1:
样例输入 2:
样例输出 2:
样例输入 3:
样例输出 3:
1
| 12 6 4 3 2 2 1 1 1 1 1 1
|
思路:
实现:
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| /*
*User: ybw051114
*Time: 2020.08.14 22:35:09
*/
#include <bits/stdc++.h>
using namespace std;
#ifndef use_ios11
#define use_ios11
using namespace std;
struct ins
{
int ans;
ins() { ans = 1; }
#define endl '\n'
void read()
{
}
void read1(char &s)
{
char c = getchar();
for (; !isprint(c) || c == ' ' || c == '\n' || c == '\t'; c = getchar())
;
s = c;
if (c == EOF)
ans = 0;
}
void read1(string &s)
{
s = "";
char c = getchar();
for (; !isprint(c) || c == ' ' || c == '\n' || c == '\t'; c = getchar())
;
for (; isprint(c) && c != ' ' && c != '\n' && c != '\t'; c = getchar())
s += c;
if (c == EOF)
ans = 0;
}
template <typename T>
void read1(T &n)
{
T x = 0;
int f = 1;
char c = getchar();
for (; !isdigit(c); c = getchar())
{
if (c == '-')
f = -1;
if (c == EOF)
{
ans = 0;
return;
}
}
for (; isdigit(c); c = getchar())
x = x * 10 + c - 48;
n = x * f;
if (c == EOF)
ans = 0;
if (c != '.')
return;
T l = 0.1;
while ((c = getchar()) <= '9' && c >= '0')
x = x + (c & 15) * l, l *= 0.1;
n = x * f;
if (c == EOF)
ans = 0;
}
void write() {}
void write1(string s)
{
int n = s.size();
for (int i = 0; i < n; i++)
putchar(s[i]);
}
void write1(const char *s)
{
int n = strlen(s);
for (int i = 0; i < n; i++)
putchar(s[i]);
}
void write1(char s) { putchar(s); }
void write1(float s, int x = 6)
{
char y[10001];
sprintf(y, "%%.%df", x);
printf(y, s);
}
void write1(double s, int x = 6)
{
char y[10001];
sprintf(y, "%%.%dlf", x);
printf(y, s);
}
void write1(long double s, int x = 6)
{
char y[10001];
sprintf(y, "%%.%dLf", x);
printf(y, s);
}
template <typename T>
void write1(T n)
{
if (n < 0)
n = -n, putchar('-');
if (n > 9)
write1(n / 10);
putchar('0' + n % 10);
}
template <typename T>
friend ins operator>>(ins x, T &n);
template <typename T>
friend ins operator<<(ins x, T n);
operator bool() { return ans; }
};
template <typename T>
ins operator>>(ins x, T &n)
{
if (!x.ans)
return x;
x.read1(n);
return x;
}
template <typename T>
ins operator<<(ins x, T n)
{
x.write1(n);
return x;
}
ins yin;
ins yout;
#endif
#include <bits/stdc++.h>
using namespace std;
const int maxn = int(1e6) + 99;
const int INF = int(1e9) + 99;
int n;
string s;
vector<int> p[2][maxn];
int nxt[2][maxn];
int ptr[2];
char buf[maxn];
int main()
{
yin >> n >> s;
for (int i = n - 1; i >= 0; --i)
{
if (s[i] != '0')
nxt[0][i] = 1 + nxt[0][i + 1];
if (s[i] != '1')
nxt[1][i] = 1 + nxt[1][i + 1];
}
for (int b = 0; b <= 1; ++b)
{
int l = 0;
while (l < n)
{
if (s[l] == char('0' + b))
{
++l;
continue;
}
int r = l + 1;
while (r < n && s[r] != char('0' + b))
++r;
for (int len = 1; len <= r - l; ++len)
p[b][len].push_back(l);
l = r;
}
}
for (int len = 1; len <= n; ++len)
{
int pos = 0, res = 0;
ptr[0] = ptr[1] = 0;
while (pos < n)
{
int npos = INF;
for (int b = 0; b <= 1; ++b)
{
if (nxt[b][pos] >= len)
npos = min(npos, pos + len);
while (ptr[b] < p[b][len].size() && pos > p[b][len][ptr[b]])
++ptr[b];
if (ptr[b] < p[b][len].size())
npos = min(npos, p[b][len][ptr[b]] + len);
}
if (npos != INF)
++res;
pos = npos;
}
yout << res << ' ';
}
yout << endl;
return 0;
}
|