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QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#748380 | #9543. Good Partitions | 4eyebird | TL | 1ms | 6372kb | C++17 | 3.7kb | 2024-11-14 20:13:46 | 2024-11-14 20:13:47 |
Judging History
answer
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <vector>
using namespace std;
typedef long long ll;
constexpr int inf = 0x3f3f3f3f;
inline char gtc()
{
static char BB[2000000], *S = BB, *T = BB;
return S == T && (T = (S = BB) + fread(BB, 1, 2000000, stdin), S == T) ? EOF : *S++;
}
typedef long long ll;
inline int read()
{
int x = 0, f = 1;
char ch = gtc();
while (ch < '0' || ch > '9')
{
if (ch == '-')
f = -1;
ch = gtc();
}
while (ch >= '0' && ch <= '9')
x = (x << 3) + (x << 1) + (ch ^ 48), ch = gtc();
return x * f;
}
inline void write(int x)
{
if (x == 0)
{
putchar('0');
return;
}
int len = 0;
static char c[25];
if (x < 0) x = -x, putchar('-');
while (x) c[len++] = x % 10 + '0', x /= 10;
while (len--) putchar(c[len]);
putchar('\n');
}
typedef vector<int> vint;
int Ns;
int stm[524288];
void update(int ps, int x, int L = 1, int R = Ns, int p = 1)
{
if (L == R)
{
stm[p] += x;
return;
}
int mid = (L + R) >> 1;
if (ps <= mid)
update(ps, x, L, mid, p << 1);
else
update(ps, x, mid + 1, R, p << 1 | 1);
stm[p] = max(stm[p << 1], stm[p << 1 | 1]);
}
int query(int L = 1, int R = Ns, int p = 1)
{
if (L == R)
return L;
int mid = (L + R) >> 1;
if (stm[p << 1 | 1] == stm[1])
return query(mid + 1, R, p << 1 | 1);
else
return query(L, mid, p << 1);
}
inline void mat(int x, int y)
{
x--;
int sp = sqrt(x);
for (int i = 1; i <= sp; i++)
{
if (x % i == 0)
{
update(i, y);
if (i != x / i)
update(x / i, y);
}
}
}
inline int js(int x)
{
int sp = sqrt(x), ret = 0;
for (int i = 1; i <= sp; i++)
{
if (x % i == 0)
{
ret++;
if (i != x / i)
ret++;
}
}
return ret;
}
int a[200010];
bool b[200010];
void solve()
{
int n = read(), q = read();
if (n == 1)
{
int x = read();
write(1);
while (q--)
{
x = read();
x = read();
write(1);
}
return;
}
Ns = n + 1;
memset(stm, 0, sizeof(stm));
for (int i = 1; i <= n; i++)
{
b[i] = 0;
a[i] = read();
}
for (int i = 2; i <= n; i++)
{
if (a[i] < a[i - 1])
{
b[i] = 1;
mat(i, 1);
}
}
if (stm[1] == 0)
write(n);
else
write(js(query()));
while (q--)
{
int p = read(), x = read();
if (b[p] && x >= a[p - 1])
{
b[p] = 0;
mat(p, -1);
}
else if (!b[p] && x < a[p - 1])
{
b[p] = 1;
mat(p, 1);
}
if (p + 1 <= n)
{
if (b[p + 1] && a[p + 1] >= x)
{
b[p + 1] = 0;
mat(p + 1, -1);
}
else if (!b[p + 1] && a[p + 1] < x)
{
b[p + 1] = 1;
mat(p + 1, 1);
}
}
a[p] = x;
if (stm[1] == 0)
write(n);
else
write(js(query()));
}
}
signed main()
{
#ifdef YGYYYGJ
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
int _T_ = read();
while (_T_--)
solve();
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 1ms
memory: 6372kb
input:
1 5 2 4 3 2 6 1 2 5 3 5
output:
1 2 3
result:
ok 3 lines
Test #2:
score: 0
Accepted
time: 1ms
memory: 5588kb
input:
1 1 1 2000000000 1 1999999999
output:
1 1
result:
ok 2 lines
Test #3:
score: -100
Time Limit Exceeded
input:
1 200000 200000 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 ...
output:
160 200000 160 200000 160 200000 160 200000 160 200000 160 200000 160 200000 160 200000 160 200000 160 200000 160 200000 160 200000 160 200000 160 200000 160 200000 160 200000 160 200000 160 200000 160 200000 160 200000 160 200000 160 200000 160 200000 160 200000 160 200000 160 200000 160 200000 160...