Write the directory title here
D. Watch the Videos
Abstract: Each video is a chain. We preset that each video needs to be viewed individually after downloading, that is, it takes 1 minute. Our goal
is to sort and divide into two legal ones (the purpose of the check function is to determine whether it is legal). The position of the long subsection, which can be combined into a chain.
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <string>
#include <math.h>
#include <vector>
#include <queue>
#include <map>
using namespace std;
void solve()
{
int n, m;
cin >> n >> m;
vector<int> a(n + 1);
long long ans = n;//每个影片都是一条链子，我们预设每个影片下载完都需要单独看一下，即单独需要1分钟
for (int i = 1; i <= n; ++ i)
{
cin >> a[i];
ans += a[i];
}
sort(a.begin() + 1, a.end());
function<int(int)> check = [&](int x)//双指针遍历这段序列是否合法
{
int l = 1, r = x;
while ( l < r)
{
if (a[l] + a[r] > m) return 0;//这个不需要判断l<r，因为进入l++后进入下一轮循环的时候while会自动判断
r --;
if (l < r && a[l] + a[r] > m) return 0;//判断r--后是否合法
l ++ ;
}
return 1;
};
int l = 1, r = n;
while(l < r)
{
int mid = l + r + 1 >> 1;
if (check(mid)) l = mid;
else r = mid - 1;
}
cout << ans - l + 1<< endl;//l包括l之前的所有影片都可以合并成一条链，但是这条链还是需要1分钟，因此最后补一个1
}
int32_t main()
{
ios::sync_with_stdio(0);
cin.tie(0);
int T = 1;
// cin >> T;
while (T --) solve();
return 0;
}
D. Range = √Sum
Push the formula. . structure. . Hey, what should I do if I don’t know the formula? I need to practice more.
Go directly to this blog , I should not write as well as him.
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <string>
#include <math.h>
#include <vector>
#include <queue>
#include <map>
using namespace std;
void solve()
{
int n;
cin >> n;
vector<int> a(n + 1);
if (n & 1)
{
for (int i = 1, j = n - n / 2 + 2; i <= n; ++ i)
{
a[i] = j ++;
}
a[1] -= 1; a[n] += 1; a[n - 1] += 1;
}
else
{
for (int i = 1, j = n - n / 2; i <= n; ++ i)
{
a[i] = j ++ ;
if (i == n / 2) j ++ ;
}
}
for (int i = 1; i <= n; ++ i) cout << a[i] << " ";
cout << endl;
}
int32_t main()
{
ios::sync_with_stdio(0);
cin.tie(0);
int T = 1;
cin >> T;
while (T --) solve();
return 0;
}