2025.1.21——1300
2025.1.21——1300
A 1300
Qingshan has a string
A string
For Div. 2 contestants, note that this condition is different from the condition in problem B.
For example,
Qingshan wants to make
- insert
Please tell Qingshan if it is possible to make
Input
The input consists of multiple test cases. The first line contains a single integer
The first line of each test case contains a single integer
The second line of each test case contains a string
It is guaranteed that
B 1300
Tema decided to improve his ice cream making skills. He has already learned how to make ice cream in a cone using exactly two balls.
Before his ice cream obsession, Tema was interested in mathematics. Therefore, he is curious about the minimum number of balls he needs to have in order to make exactly
There are plenty possible ice cream flavours:
Two ice creams are considered different if their sets of ball flavours are different. For example,
For example, having the following ice cream balls:
Note, that Tema do not need to make all the ice cream cones at the same time. This means that he making ice cream cones independently. Also in order to make a following cone
Help Tema answer this question. It can be shown that answer always exist.
Input
Each test consists of multiple test cases. The first line of input contains a single integer
The first line of each test case contains a single integer
------------------------思考------------------------
-
字符串/贪心策略+思维/贪心策略/二分/数学
A
- 首先保证
- 只能在两边加串，贪心消除即可，操作次数至多
B
- 是道好题。二分找出上界，剩余部分贪心，思想上证明一下。
------------------------代码------------------------
A
#include <bits/stdc++.h>
#define int long long //
#define endl '\n' // attention: interactive/debug
#define el cout << endl
using namespace std;
#define bug(BUG) cout << "bug:# " << (BUG) << endl
#define bug2(BUG1, BUG2) cout << "bug:# " << (BUG1) << " " << (BUG2) << endl
#define bug3(BUG1, BUG2, BUG3) cout << "bug:# " << (BUG1) << ' ' << (BUG2) << ' ' << (BUG3) << endl
#define bugv(VEC, ST) \
for (int i = ST; i < VEC.size(); i++) \
cout << VEC[i] << ' '; \
el;
void _();
signed main()
{
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cout << fixed << setprecision(10);
int T = 1;
cin >> T;
while (T--)
_();
return 0;
}
void _()
{
int n;
cin >> n;
string s;
cin >> s;
int cnt[2] = {};
for (auto v : s)
cnt[v - '0']++;
if (cnt[0] - cnt[1])
{
cout << -1;
el;
return;
}
int l = 0, r = n - 1;
vector<int> res;
// int t = 1e3;
while (l < r)
{
// t--;
// if (!t)
// break;
if (s[l] - s[r])
{
l++, r--;
continue;
}
if (s[l] == '1')
{
res.push_back(l);
s = s.substr(0, l) + "01" + s.substr(l);
}
else
{
res.push_back(r + 1);
// bug2(s.substr(0, r + 1), s.substr(r + 1));
s = s.substr(0, r + 1) + "01" + s.substr(r + 1);
}
l++, r++;
// bug2(l, r);
}
cout << res.size();
el;
bugv(res, 0);
}
B
#include <bits/stdc++.h>
#define int long long //
#define endl '\n' // attention: interactive/debug
#define el cout << endl
using namespace std;
#define bug(BUG) cout << "bug:# " << (BUG) << endl
#define bug2(BUG1, BUG2) cout << "bug:# " << (BUG1) << " " << (BUG2) << endl
#define bug3(BUG1, BUG2, BUG3) cout << "bug:# " << (BUG1) << ' ' << (BUG2) << ' ' << (BUG3) << endl
#define bugv(VEC, ST) \
{ \
for (int I = ST; I < VEC.size(); I++) \
cout << VEC[I] << ' '; \
el; \
}
void _();
signed main()
{
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cout << fixed << setprecision(10);
int T = 1;
cin >> T;
while (T--)
_();
return 0;
}
void _()
{
int n;
cin >> n;
int l = 1, r = 1e10;
while (r - l - 1)
{
int mid = l + r >> 1;
if (mid * (mid - 1) >> 1 > n)
r = mid;
else
l = mid;
}
int res = l;
res += n - (l * (l - 1) >> 1);
cout << res << endl;
}
// void _()
// {
// auto cal = [](int mid)
// {
// return mid * (mid - 1) >> 1;
// };
// for (int i = 2; i <= 20; i++)
// bug2(i, cal(i));
// }
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