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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 |
---|---|---|---|---|---|---|---|---|---|
#811287 | #241. Chiaki Sequence Revisited | rlc202204 | 100 ✓ | 96ms | 3828kb | C++17 | 2.1kb | 2024-12-12 17:26:00 | 2024-12-12 17:26:01 |
Judging History
answer
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 105;
const int mod = 1e9 + 7;
void add(int &x, int y) {
x = (x + y) % mod;
}
int a[N] = {0};
void fnd() {
int n;
cin >> n;
a[1] = 1, a[2] = 1;
for (int i = 3; i <= n; i++)
a[i] = a[i - a[i - 1]] + a[i - 1 - a[i - 2]];
int l = 2, r;
while (l <= n) {
r = l;
int cnt = 0;
while (r <= n && a[r] == a[l])
cnt++, r++;
printf("(%d, %d): %d cnt = %d\n", l - 1, r - 2, a[l], cnt);
l = r;
}
}
int pw[100] = {0};
int pws[100] = {0};
//?? lowbit ??
int res[100] = {0};
long long cnt[100] = {0};
int cal(int k) {
//?? 1 ~ 2^k - 1 ? i log(lowbit(i))
int ans = 0;
for (int j = 1; j <= k; j++) {
int sum = 1ll * pws[k - j] * pw[j] % mod;
add(sum, 1ll * pw[j - 1] * pw[k - j] % mod);
add(ans, 1ll * j * sum % mod);
cnt[k] += j * (1ll << (k - j));
// if (k == 2)
// cout << j << " " << sum << endl;
}
return ans;
}
int f(long long n, int k) {
if (n == 0ll)
return 0;
// cout << "At " << n << " " << k << " " << endl;
if (k == 1)
return 1;
if (n <= cnt[k - 1])
return f(n, k - 1);
if (n - cnt[k - 1] <= k)
return (res[k - 1] + (n - cnt[k - 1]) * pw[k - 1] % mod) % mod;
int ans = 0;
add(ans, res[k - 1]);
add(ans, 1ll * k * pw[k - 1] % mod);
// cout << k << " : " << "ans: " << ans << endl;
n -= cnt[k - 1];
n -= k;
add(ans, n % mod * pw[k - 1] % mod);
add(ans, f(n, k - 1));
return ans;
}
void slv() {
long long n;
cin >> n;
n--;
cout << (f(n, 60) + 1) % mod << endl;
}
int main() {
// fnd();
pw[0] = 1;
for (int i = 1; i <= 60; i++)
pw[i] = 2ll * pw[i - 1] % mod;
for (int i = 1; i <= 60; i++)
pws[i] = 1ll * (pw[i] - 1) * pw[i - 1] % mod;
for (int i = 0; i <= 60; i++) {
res[i] = cal(i);
// if (i <= 5)
// cout << i << " " << cnt[i] << endl;
/* if (i < 5) {
int ans = 0;
for (int j = 1; j < (1 << i); j++) {
int k = 0;
while (!(j >> k & 1))
k++;
ans += (k + 1) * j;
}
// cout << "Chk " << ans << endl;
}*/
}
// return 0;
int T;
scanf("%d", &T);
while (T--)
slv();
return 0;
}
詳細信息
Pretests
Final Tests
Test #1:
score: 100
Accepted
time: 96ms
memory: 3828kb
input:
100000 1 2 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 99 100 1...
output:
1 2 4 6 9 13 17 21 26 32 38 45 53 61 69 77 86 96 106 117 129 141 153 166 180 194 209 225 241 257 273 289 306 324 342 361 381 401 421 442 464 486 509 533 557 581 605 630 656 682 709 737 765 793 822 852 882 913 945 977 1009 1041 1073 1105 1138 1172 1206 1241 1277 1313 1349 1386 1424 1462 1501 1541 158...
result:
ok 100000 lines