AtCoder Regular Contest 082 ABCD

 A

#include<bits/stdc++.h>
using namespace std;
int a[123456];
int n,m;
int main(){
    cin>>n>>m;
    cout<<max(0,n-m)<<endl;
    return 0;
}

B

#include<bits/stdc++.h>
using namespace std;
int a[123456];
int n,m;
int main(){
    string s;
    cin>>s;
    string ans = "";
    for (int i = 0; i < s.length(); i += 2)
    {
        ans += s[i];
    }
    cout << ans << endl;
    return 0;
}

Problem Statement

You are given an integer sequence of length N, a1,a2,…,aN.

For each 1≤i≤N, you have three choices: add 1 to ai, subtract 1 from ai or do nothing.

After these operations, you select an integer X and count the number of i such that ai=X.

Maximize this count by making optimal choices.

Constraints

• 1≤N≤10⁵
• 0≤ai<10⁵(1≤i≤N)
• ai is an integer.

---

Input

The input is given from Standard Input in the following format:

N
a1 a2 .. aN

Output

Print the maximum possible number of i such that ai=X.

---

Sample Input 1

Copy

7
3 1 4 1 5 9 2

Sample Output 1

Copy

4

For example, turn the sequence into 2,2,3,2,6,9,2 and select X=2 to obtain 4, the maximum possible count.

---

Sample Input 2

Copy

10
0 1 2 3 4 5 6 7 8 9

Sample Output 2

Copy

3

---

Sample Input 3

Copy

1
99999

Sample Output 3

Copy

1

解法：数字-1 +1 或者不变，最后留下数字最多的次数
解法：那就...讨论一下
本身的出现次数，i和i+1 i和i+2 i和i+1和i+2的出现次数

#include<bits/stdc++.h>
using namespace std;
long long a[123456];
int n;
bool vis[1009];
map<long long,long long>Mp;
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        Mp[a[i]]++;
    }
    long long Max=1;
    for(int i=0;i<=100010;i++){
            Max=max(Max,Mp[i]);
            if(Mp[i]&&Mp[i+1]&&Mp[i+2]){
               long long ans=Mp[i]+Mp[i+1]+Mp[i+2];
               Max=max(Max,ans);
            }
            if(Mp[i]&&Mp[i+1]){
               long long ans=Mp[i]+Mp[i+1];
               Max=max(Max,ans);
            }
            if(Mp[i]&&Mp[i+2]){
               long long ans=Mp[i]+Mp[i+2];
               Max=max(Max,ans);
            }
    }
    cout<<Max<<endl;
    return 0;
}

D - Derangement

---

Time limit : 2sec / Memory limit : 256MB

Score : 400 points

Problem Statement

You are given a permutation p1,p2,…,pN consisting of 1,2,..,N. You can perform the following operation any number of times (possibly zero):

Operation: Swap two adjacent elements in the permutation.

You want to have pi≠i for all 1≤i≤N. Find the minimum required number of operations to achieve this.

Constraints

• 2≤N≤10⁵
• p1,p2,..,pN is a permutation of 1,2,..,N.

---

Input

The input is given from Standard Input in the following format:

N
p1 p2 .. pN

Output

Print the minimum required number of operations

---

Sample Input 1

Copy

5
1 4 3 5 2

Sample Output 1

Copy

2

Swap 1 and 4, then swap 1 and 3. p is now 4,3,1,5,2 and satisfies the condition. This is the minimum possible number, so the answer is 2.

---

Sample Input 2

Copy

2
1 2

Sample Output 2

Copy

1

Swapping 1 and 2 satisfies the condition.

---

Sample Input 3

Copy

2
2 1

Sample Output 3

Copy

0

The condition is already satisfied initially.

---

Sample Input 4

Copy

9
1 2 4 9 5 8 7 3 6

Sample Output 4

Copy

3

题意：交换相邻的数字，使得ai!=i 问最少的次数

解法：贪心，反正如果是正确位置上的，我们去交换相邻的就可以了

#include<bits/stdc++.h>
using namespace std;
int a[123456];
int n;
bool vis[1009];
map<int,int>Mp;
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        if(a[i]==i){
            Mp[i]=1;
        }
    }
    int sum=0;
    for(int i=1;i<=n;i++){
        if(Mp[i]==1){
            sum++;
            Mp[i]=0;
            Mp[i+1]=0;
        }
    }
    cout<<sum<<endl;
    return 0;
}