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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 |
|---|---|---|---|---|---|---|---|---|---|
| #459613 | #8834. Formal Fring | ucup-team1231# | AC ✓ | 90ms | 11740kb | C++14 | 763b | 2024-06-30 09:37:17 | 2024-06-30 09:40:00 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
int n, dp[1000005], fdp[1000005];
int main() {
scanf("%d", &n);
dp[0] = 1;
for(int i = 1; i <= n; i <<= 1)
for(int j = i; j <= n; j++)
dp[j] = (dp[j] + dp[j - i]) % MOD;
fdp[0] = 1;
for(int i = 1; i <= n; i++) {
for(int j = 0; i >> j; j++) {
int c = i >> j;
if((c + 1) & c) continue;
int t = i ^ c << j;
fdp[i] = (fdp[i] + 1LL * dp[c] * ((t << 1) >> j ? fdp[i ^ c << j] : dp[i ^ c << j])) % MOD;
}
fdp[i] = (dp[i] - fdp[i] + MOD) % MOD;
}
for(int i = 1; i <= n; i++)
printf("%d ", (dp[i] - fdp[i] + MOD) % MOD);
return 0;
}
详细
Test #1:
score: 100
Accepted
time: 1ms
memory: 6020kb
input:
10
output:
1 1 2 1 1 3 6 1 1 2
result:
ok 10 numbers
Test #2:
score: 0
Accepted
time: 1ms
memory: 6032kb
input:
70
output:
1 1 2 1 1 3 6 1 1 2 2 5 5 11 26 1 1 2 2 4 4 6 6 11 11 16 16 27 27 53 166 1 1 2 2 4 4 6 6 10 10 14 14 20 20 26 26 37 37 48 48 64 64 80 80 107 107 134 134 187 187 353 1626 1 1 2 2 4 4 6
result:
ok 70 numbers
Test #3:
score: 0
Accepted
time: 90ms
memory: 11740kb
input:
1000000
output:
1 1 2 1 1 3 6 1 1 2 2 5 5 11 26 1 1 2 2 4 4 6 6 11 11 16 16 27 27 53 166 1 1 2 2 4 4 6 6 10 10 14 14 20 20 26 26 37 37 48 48 64 64 80 80 107 107 134 134 187 187 353 1626 1 1 2 2 4 4 6 6 10 10 14 14 20 20 26 26 36 36 46 46 60 60 74 74 94 94 114 114 140 140 166 166 203 203 240 240 288 288 336 336 400 ...
result:
ok 1000000 numbers
Extra Test:
score: 0
Extra Test Passed