标签：752 sequence int ll Codeforces long cin Div define
contents:
A. Era
Shohag has an integer sequence a1,a2,…,an. He can perform the following operation any number of times (possibly, zero):
Select any positive integer k (it can be different in different operations).
Choose any position in the sequence (possibly the beginning or end of the sequence, or in between any two elements) and insert k into the sequence at this position.
This way, the sequence a changes, and the next operation is performed on this changed sequence.
For example, if a=[3,3,4] and he selects k=2, then after the operation he can obtain one of the sequences [2–,3,3,4], [3,2–,3,4], [3,3,2–,4], or [3,3,4,2–].
Shohag wants this sequence to satisfy the following condition: for each 1≤i≤|a|, ai≤i. Here, |a| denotes the size of a.
Help him to find the minimum number of operations that he has to perform to achieve this goal. We can show that under the constraints of the problem it’s always possible to achieve this goal in a finite number of operations.
AC
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define db double
#define pii pair<int, int>
#define psi pair<string, int>
#define ull unsigned ll
#define pb push_back
#define mp make_pair
#define X first
#define Y second
#define ld long double
const int N = 1E2 + 7;
#define INF ~0ULL
int t;
int n;
int arr[N];
int main()
{
cin >> t;
while (t--)
{
cin >> n;
for (int i = 1; i <= n; i++)
{
cin >> arr[i];
}
int ans = 0;
for (int i = 1; i <= n; i++)
{
ans = max (arr[i]-i,ans);
}
cout<<ans<<endl;
}
}
B. XOR Specia-LIS-t
YouKn0wWho has an integer sequence a1,a2,…an. Now he will split the sequence a into one or more consecutive subarrays so that each element of a belongs to exactly one subarray. Let k be the number of resulting subarrays, and h1,h2,…,hk be the lengths of the longest increasing subsequences of corresponding subarrays.
For example, if we split [2,5,3,1,4,3,2,2,5,1] into [2,5,3,1,4], [3,2,2,5], [1], then h=[3,2,1].
YouKn0wWho wonders if it is possible to split the sequence a in such a way that the bitwise XOR of h1,h2,…,hk is equal to 0. You have to tell whether it is possible.
The longest increasing subsequence (LIS) of a sequence b1,b2,…,bm is the longest sequence of valid indices i1,i2,…,ik such that i1<i2<…<ik and bi1<bi2<…<bik. For example, the LIS of [2,5,3,3,5] is [2,3,5], which has length 3.
An array c is a subarray of an array b if c can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
思路
偶数直接过
奇数找一对凑成偶数
如果连二都凑不出来，只能说明原数组的最长
递增子序列为n，奇数个的xor结果不会为0
AC
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define db double
#define pii pair<int, int>
#define psi pair<string, int>
#define ull unsigned ll
#define pb push_back
#define mp make_pair
#define X first
#define Y second
#define ld long double
const int N = 1E5 + 7;
#define INF ~0ULL
int t;
int n;
int arr[N];
int main()
{
cin >> t;
while (t--)
{
cin >> n;
for (int i = 1; i <= n; i++)
{
cin >> arr[i];
}
bool flag = 0;
if (n % 2 == 0)
{
cout << "YES" << endl;
}
else
{
for (int i = 2; i <= n; i++)
{
if (arr[i] <= arr[i - 1])
{
flag = 1;
break;
}
}
if (flag)
cout << "YES" << endl;
else
cout << "NO" << endl;
}
}
}
C. Di-visible Confusion
YouKn0wWho has an integer sequence a1,a2,…,an. He will perform the following operation until the sequence becomes empty: select an index i such that 1≤i≤|a| and ai is not divisible by (i+1), and erase this element from the sequence. Here |a| is the length of sequence a at the moment of operation. Note that the sequence a changes and the next operation is performed on this changed sequence.
For example, if a=[3,5,4,5], then he can select i=2, because a2=5 is not divisible by i+1=3. After this operation the sequence is [3,4,5].
Help YouKn0wWho determine if it is possible to erase the whole sequence using the aforementioned operation.
思路
模拟
AC
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define db double
#define pii pair<int, int>
#define psi pair<string, int>
#define ull unsigned ll
#define pb push_back
#define mp make_pair
#define X first
#define Y second
#define ld long double
const int N = 1E5 + 7;
#define INF ~0ULL
int t;
int n;
int arr[N];
bool cc()
{
cin >> n;
for (int i = 1; i <= n; i++)
{
cin >> arr[i];
}
for (int i = 1; i <= n; i++)
{
bool f = 1;
for (int j = i; j >= 1; j--)
{
if (arr[i] % (j + 1))
{
f = 0;
break;
}
}
if (f)
return false;
}
return true;
}
int main()
{
cin >> t;
while (t--)
{
if (cc()) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
}
D. Moderate Modular Mode
YouKn0wWho has two even integers x and y. Help him to find an integer n such that 1≤n≤2⋅1018 and nmodx=ymodn. Here, amodb denotes the remainder of a after division by b. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints.
思路
找个值使得n%x,y%n
如果x=y n=x
x>y 凑成x=y (mody) ans = x*y+y
x<y y - y % x / 2
AC
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void cc()
{
ll a, b;
cin >> a >> b;
if (a > b)
{
ll res;
res = a * b + b;
cout << res << endl;
}
else if (a == b)
{
cout << a << endl;
}
else
{
cout << b - b % a / 2 << endl;
}
}
int main()
{
int t;
cin >> t;
while (t--)
{
cc();
}
return 0;
}
标签：752,sequence,int,ll,Codeforces,long,cin,Div,define 来源： https://blog.csdn.net/MoRan0_0/article/details/121063957
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