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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#759463 | #9553. The Hermit | mango09 | TL | 0ms | 4064kb | C++14 | 2.0kb | 2024-11-18 09:01:48 | 2024-11-18 09:01:50 |
Judging History
answer
// Ice cream? I scream!
#include <bits/stdc++.h>
#ifdef ONLINE_JUDGE
#define freopen Chitanda
#endif
#define wjy namespace
#define xsy std
typedef long long ll;
using wjy xsy;
template<typename T> void debug(string s, T x) {cerr << " " << s << " = " << x << " \n";}
template<typename T, typename... Args> void debug(string s, T x, Args... args) {for (int i = 0, b = 0; i < (int)s.size(); i++) if (s[i] == '(' || s[i] == '{') b++;else if (s[i] == ')' || s[i] == '}') b--;else if (s[i] == ',' && b == 0) {cerr << " " << s.substr(0, i) << " = " << x << " |";debug(s.substr(s.find_first_not_of(' ', i + 1)), args...);break;}}
#ifdef ONLINE_JUDGE
#define Debug(...)
#else
#define Debug(...) debug(#__VA_ARGS__, __VA_ARGS__)
#endif
const int MAXN = 1e5 + 5;
const int MOD = 998244353;
int Qpow(int a, int b) {int base = a, ans = 1; while (b) {if (b & 1) ans = 1ll * ans * base % MOD; base = 1ll * base * base % MOD; b >>= 1;} return ans;}
int Inv(int a) {return Qpow(a, MOD - 2);}
int fac[MAXN], ifac[MAXN];
void Init(int n)
{
fac[0] = 1;
for (int i = 1; i <= n; i++)
fac[i] = 1ll * fac[i - 1] * i % MOD;
ifac[n] = Inv(fac[n]);
for (int i = n - 1; ~i; i--)
ifac[i] = 1ll * ifac[i + 1] * (i + 1) % MOD;
}
int C(int n, int m)
{
if (n < m)
return 0;
return 1ll * fac[n] * ifac[n - m] % MOD * ifac[m] % MOD;
}
int m, n, cnt;
int d[MAXN];
map<pair<int, int>, int> f;
int main()
{
scanf("%d%d", &m, &n);
Init(m);
int ans = 1ll * n * C(m, n) % MOD, tot = 0;
// Debug(C(m, n), ans);
for (int i = 1; i <= m; i++)
{
f[{i, 1}] = 1;
for (int k = 2; i * k <= m; k++)
for (int j = 1; j <= ceil(log2(i)) + 1; j++)
f[{i * k, j + 1}] = (f[{i * k, j + 1}] + f[{i, j}]) % MOD;
}
for (int i = 1; i <= m; i++)
{
for (int j = 1; j <= ceil(log2(i)) + 1; j++)
{
// Debug(i, j, f[{i, j}]);
tot = (tot + 1ll * f[{i, j}] * C((m / i) - 1, n - j) % MOD) % MOD;
}
}
// Debug(tot);
// tot = (tot + 1ll * (n - 1) * cnt % MOD) % MOD;
printf("%d\n", (ans - tot + MOD) % MOD);
return 0;
}
Details
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Test #1:
score: 100
Accepted
time: 0ms
memory: 4064kb
input:
4 3
output:
7
result:
ok 1 number(s): "7"
Test #2:
score: 0
Accepted
time: 0ms
memory: 4044kb
input:
11 4
output:
1187
result:
ok 1 number(s): "1187"
Test #3:
score: -100
Time Limit Exceeded
input:
100000 99999