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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 |
---|---|---|---|---|---|---|---|---|---|
#572977 | #2519. Number with Bachelors | lifan | AC ✓ | 235ms | 26412kb | C++14 | 3.0kb | 2024-09-18 16:58:41 | 2024-09-18 16:58:42 |
Judging History
answer
//
//什么时候ICPC的题也开始“重视”输入输出了啊。。。。。。。
//你非要求两个进制的答案才舒服吗。。。。
#include <bits/stdc++.h>
using namespace std;
const int maxn = (1 << 16) + 10;
typedef unsigned long long ull;
ull dp[2][22][maxn];
int typ, a[100];
ull dfs(int x, int sta, int limit)
{
if (x == -1) return 1;
int ty = (typ == 10) ? 0 : 1;
if (dp[ty][x][sta] != -1 && !limit) return dp[ty][x][sta];
int up = limit ? a[x] : typ - 1;
ull ans = 0;
for (int i = 0; i <= up; i++)
{
if ((sta >> i) & 1) continue;
if (sta == 0 && i == 0) ans += dfs(x - 1, sta, limit && i == up);
else ans += dfs(x - 1, sta | (1 << i), limit && i == up);
}
if (!limit) dp[ty][x][sta] = ans;
return ans;
}
ull solve(ull x)
{
int num = 0;
while (x)
{
a[num++] = x % typ;
x /= typ;
}
return dfs(num - 1, 0, true);
}
void input(ull &x)
{
x = 0;
char s[22];
scanf("%s", s + 1);
int len = strlen(s + 1);
for (int i = 1; i <= len; i++)
{
if (s[i] <= '9' && s[i] >= '0') x = x * typ + s[i] - '0';
else x = x * typ + s[i] - 'a' + 10;
}
}
void print(ull x)
{
if (x == 0)
{
printf("0\n");
return;
}
if (typ == 10) printf("%llu\n", x);
else
{
vector<int> ans;
while (x)
{
int p = x % typ;
x /= typ;
ans.push_back(p);
}
for (int i = ans.size() - 1; i >= 0; i--)
{
if (ans[i] >= 10) printf("%c", ans[i] - 10 + 'a');
else printf("%c", ans[i] + '0');
}
printf("\n");
}
}
void test()
{
int t;
scanf("%d", &t);
while (t--)
{
int x;
scanf("%d%d", &x, &typ);
printf("%llu", solve(x));
}
}
int main()
{
memset(dp, -1, sizeof(dp));
int T;
scanf("%d", &T);
while (T--)
{
char op[10];
scanf("%s", op);
if (op[0] == 'd') typ = 10;
else typ = 16;
int flg;
scanf("%d", &flg);
if (!flg)
{
ull a, b;
input(a), input(b);
ull ans = solve(b);
if (a > 0) ans -= solve(a - 1);
print(ans);
}
else
{
ull x;
input(x);
if (x < 10)
{
printf("%llu\n", x - 1);
continue;
}
ull l = 0, r = 0, ans = 0;
r--;
if (solve(r) < x)
{
printf("-\n");
continue;
}
while (l <= r)
{
ull mid = l + (r - l) / 2;
if (solve(mid) >= x) r = mid - 1, ans = mid;
else l = mid + 1;
}
print(ans);
}
}
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 42ms
memory: 26412kb
input:
6 d 0 10 20 h 0 10 1f d 1 10 h 1 f d 1 1000000000 h 1 ffffffffffffffff
output:
10 f 9 e - -
result:
ok 6 lines
Test #2:
score: 0
Accepted
time: 235ms
memory: 26336kb
input:
50000 h 1 147a d 0 25 71 d 1 3587 d 0 26922 51887 d 1 711 d 0 3 5 h 0 7bf2defab442a0b1 f299a4cf1d4d9bed d 0 6961 91018 d 1 4 d 1 876 h 1 12cc5d3370f99120 d 1 161315 h 0 25f 6959 d 0 467 516 d 1 298 h 1 70260cdc2da73281 h 1 928e17d65d764ca2 h 1 c8ec8a7b67605e51 d 1 91697 d 0 4941925161850941148 89850...
output:
1b36 43 6587 7710 953 3 8daab378500 26054 3 1356 - 946307 4681 40 387 - - - 491850 0 1 29 - 4605298 1 1 - 15b4 175f 9b944134000 124b7 6279 9 6257 - 39be22a900 5c636b59300 146ce 2a55 - 0 - 7 d 6 2041 - 1c94afe7300 0 5 9149 16540973 1389 125 0 - 3bc31189480 424 66800 7 - - 1e6 0 0 48b6 9 - 2b0 5019 14...
result:
ok 50000 lines