AtCoder Regular Contest 082 ABCD

 A

1 #include<bits/stdc++.h>
2 using namespace std;
3 int a[123456];
4 int n,m;
5 int main(){
6     cin>>n>>m;
7     cout<<max(0,n-m)<<endl;
8     return 0;
9 }

B

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int a[123456];
 4 int n,m;
 5 int main(){
 6     string s;
 7     cin>>s;
 8     string ans = "";
 9     for (int i = 0; i < s.length(); i += 2)
10     {
11         ans += s[i];
12     }
13     cout << ans << endl;
14     return 0;
15 }

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
 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 long long a[123456];
 4 int n;
 5 bool vis[1009];
 6 map<long long,long long>Mp;
 7 int main(){
 8     scanf("%d",&n);
 9     for(int i=1;i<=n;i++){
10         scanf("%d",&a[i]);
11         Mp[a[i]]++;
12     }
13     long long Max=1;
14     for(int i=0;i<=100010;i++){
15             Max=max(Max,Mp[i]);
16             if(Mp[i]&&Mp[i+1]&&Mp[i+2]){
17                long long ans=Mp[i]+Mp[i+1]+Mp[i+2];
18                Max=max(Max,ans);
19             }
20             if(Mp[i]&&Mp[i+1]){
21                long long ans=Mp[i]+Mp[i+1];
22                Max=max(Max,ans);
23             }
24             if(Mp[i]&&Mp[i+2]){
25                long long ans=Mp[i]+Mp[i+2];
26                Max=max(Max,ans);
27             }
28     }
29     cout<<Max<<endl;
30     return 0;
31 }

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;
}

 

posted @ 2017-09-02 23:36  樱花落舞  阅读(1164)  评论(0)  编辑  收藏  举报