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A-尖子生
分成jzc、jz、zc、d四种情况即可
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| #include <iostream>
#include <cstring>
using namespace std;
int main()
{
string s;
cin >> s;
int c1 = 0, c2 = 0, c3 = 0;
for(int i = 0; i < s.size(); i++)
{
if(s[i] == 'j')
{
if(s[i + 1] == 'z')
{
if(s[i + 2] == 'c')
{
c1++, c2++;
i += 2;
}
else
{
c1++;
i += 1;
}
}
else
{
continue;
}
}
else if(s[i] == 'z')
{
if(s[i + 1] == 'c')
{
c2++;
i += 1;
}
else
continue;
}
else if(s[i] == 'd')
{
c3++;
}
}
cout << c1 << " " << c2 << " " << c3 ;
return 0;
}
|
B
B-建房子
考察离散化，用结构体存下搭建的起点和终点，起点标记为1，终点标记为-1，通过加上起点或终点的标记判断当前的高度，为保证正确性，起点应该排在终点的前面。如果当前高度大于已经记录的高度，就更新高度和最高的个数；如果当前高度等于最高高度，就加上最高高度
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| #include <iostream>
#include <algorithm>
using namespace std;
#define int long long
const int N = 1e6 + 10;
struct re
{
int x, f;
}a[N * 2];
bool cmp(re a, re b)
{
if(a.x == b.x) return a.f > b.f;
return a.x < b.x;
}
signed main()
{
int n;
cin >> n;
for(int i = 1; i <= n; i++)
{
cin >> a[i * 2 - 1].x, a[i * 2 - 1].f = 1;
cin >> a[i * 2].x, a[i * 2].f = -1;
}
int nh = 0;
int mc = 0, mh = 0;
for(int i = 1; i <= 2 * n; i++)
{
nh += a[i].f;
if(nh > mh)
{
mh = nh;
mc = a[i + 1].x - a[i].x + 1;
}
else if(nh == mh)
{
mc += a[i + 1].x - a[i].x + 1;
}
}
cout << mh << " " << mc;
return 0;
}
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C
C-连通图
分成环、不成环；o可以1:2连接，也2:2连接
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| #include <iostream>
using namespace std;
#define int long long
void solve()
{
int c, o, h;
cin >> c >> o >> h;
if(h % 2 == 1 || h > 2 * c + 2)
{
cout << "NO" << '\n';
return;
}
if(c == 0)
{
if(h == 2) cout << "YES" << '\n';
else if(o > 2 && h == 0) cout << "YES" << '\n';
else cout << "NO" << '\n';
}
else if(c == 1)
{
if(o >= 2 * c + 2 - h) cout << "YES" << '\n';
else cout << "NO" << '\n';
}
else if(c == 2)
{
if(o >= (2 * c + 2 - h) / 2) cout << "YES" << '\n';
else cout << "NO" << '\n';
}
else
{
if(o >= (2 * c + 2 - h) / 2) cout << "YES" << '\n';
else if(h <= 2 * c)
{
if(o >= (2 * c - h) / 2) cout << "YES" << '\n';
else cout << "NO" << '\n';
}
else cout << "NO" << '\n';
}
}
signed main()
{
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
int t;
cin >> t;
while(t--)
{
solve();
}
return 0;
}
|
10-16
A
A-Happy!
美咲的翻翻乐
模拟即可
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| #include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int a[N];
int ans = 0;
int main()
{
int n;
cin >> n;
int c = 0;
n = 2 * n;
while(n--)
{
int x;
cin >> x;
if(!a[x])
{
a[x] = 1;
c++;
ans = max(ans, c);
}
else
c--;
}
cout << ans;
return 0;
}
|
B
B-Lucky!
米歇尔的加法
先将两个数补齐，转换成十进制，然后这个“加法”相当于异或，再转换成十六进制，删除前导零即可。注意如果是全零时的特判
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| #include <iostream>
#include <vector>
using namespace std;
vector<char> a;
int main()
{
string x0, y;
cin >> x0 >> y;
string x = "";
if (x0.size() > y.size()) swap(x0, y);
for (int i = y.size() - x0.size(); i > 0; i--)
{
x += '0';
}
x += x0;
for (int i = 0; i < x.size(); i++)
{
int x10, y10;
if (x[i] >= 'A' && x[i] <= 'Z')
x10 = x[i] - 'A' + 10;
else
x10 = x[i] - '0';
if (y[i] >= 'A' && y[i] <= 'Z')
y10 = y[i] - 'A' + 10;
else
y10 = y[i] - '0';
int ians = x10 ^ y10;
char cans;
if (ians >= 0 && ians <= 9) cans = ians + '0';
if (ians >= 10 && ians <= 15) cans = ians - 10 + 'A';
a.push_back(cans);
}
int i = 0;
while (i < a.size() && a[i] == '0') i++;
if (i == a.size()) cout << 0;
for (; i < a.size(); i++)
cout << a[i];
return 0;
}
|
C
C-Smile!
花音的礼物分配
利用背包问题，列举所有可能的情况
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| #include <bits/stdc++.h>
using namespace std;
vector<int> ac;
const int N = 510;
int sum1 = 0, sum2 = 0, s = 0, f[N * N];
int main()
{
int n;
cin >> n;
while(n --)
{
int a, b;
cin >> a >> b;
if(b == 0) ac.push_back(a), s += a;
else if(b == 1) sum1 += a;
else sum2 += a;
}
int ans = s + sum1 + sum2;
f[sum1] = 1;
for(int i = 0; i < ac.size(); i++)
{
for(int j = sum1 + s; j >= ac[i]; j--)
{
f[j] |= f[j - ac[i]];
}
}
for(int i = 0; i <= s + sum1; i++)
{
if(f[i])
{
ans = min(ans, abs(s + sum1 + sum2 - 2 * i));
}
}
cout << ans;
return 0;
}
|
D
D-Yeah!!
弦卷心的好子串
双指针法，记录h、hw、w的个数
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| #include <bits/stdc++.h>
#define int long long
using namespace std;
int n, k;
string x, a = "";
int h, w, hw, s, ans;
signed main()
{
cin >> n >> k;
cin >> x;
for(int i = 0; i < n; i++)
if(x[i] == 'h' || x[i] == 'w') a += x[i];
int m = a.size();
for(int i = 0, j = -1; i < m; i++)
{
while(j < m && s < k)
{
j++;
if(a[j] == 'h') h++;
else
{
w++;
hw += h;
s += h * (h - 1) / 2;
}
}
ans += m - j;
if(a[i] == 'h')
{
h--;
hw -= w;
s -= hw;
}
else
w--;
}
cout << ans;
return 0;
}
|
