AtCoder Beginner Contest 072

A - Sandglass2

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Time limit : 2sec / Memory limit : 256MB

Score : 100 points

Problem Statement

We have a sandglass that runs for X seconds. The sand drops from the upper bulb at a rate of 1 gram per second. That is, the upper bulb initially contains X grams of sand.

How many grams of sand will the upper bulb contains after t seconds?

Constraints

• 1≤X≤109
• 1≤t≤109
• X and t are integers.

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Input

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

X t

Output

Print the number of sand in the upper bulb after t second.

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Sample Input 1

100 17

Sample Output 1

83

17 out of the initial 100 grams of sand will be consumed, resulting in 83 grams.

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Sample Input 2

48 58

Sample Output 2

0

All 48 grams of sand will be gone, resulting in 0 grams.

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Sample Input 3

1000000000 1000000000

Sample Output 3

0

#include<iostream>
using namespace std;
int main()
{
	long long int x,t;
	cin>>x>>t;
	if(x-t==0)
	{
		cout<<0<<endl;
	}
	if(x-t>0)
	{
		cout<<x-t<<endl;
	}
	if(x-t<0)
	{
		cout<<0<<endl;
	}
	return 0;
}

B - OddString

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Time limit : 2sec / Memory limit : 256MB

Score : 200 points

Problem Statement

You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.

Constraints

• Each character in s is a lowercase English letter.
• 1≤|s|≤105

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Input

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

s

Output

Print the string obtained by concatenating all the characters in the odd-numbered positions.

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Sample Input 1

atcoder

Sample Output 1

acdr

Extract the first character a, the third character c, the fifth character d and the seventh character r to obtain acdr.

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Sample Input 2

aaaa

Sample Output 2

aa

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Sample Input 3

z

Sample Output 3

z

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Sample Input 4

fukuokayamaguchi

Sample Output 4

fkoaaauh

#include<iostream>
using namespace std;
char a[100010];
int main()
{
	long long int n=0;
	while(~scanf("%c",&a[n]))
	{
		n++;
	}
	for(int i=0;i<n;i=i+2)
		{
			cout<<a[i];
		}
	return 0;
}

C - Together

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Time limit : 2sec / Memory limit : 256MB

Score : 300 points

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.

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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.

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Sample Input 1

7
3 1 4 1 5 9 2

Sample Output 1

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.

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Sample Input 2

10
0 1 2 3 4 5 6 7 8 9

Sample Output 2

3

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Sample Input 3

1
99999

Sample Output 3

1

#include <iostream>
#include<cstring>
#include <algorithm>
using namespace std;
int a[100001];
int main()
{
    int n;
    while(cin>>n)
    {
        int t;
        memset(a,0,sizeof(a));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&t);
            a[t+1]++;
            a[t-1]++;
            a[t]++;
        }
        sort(a,a+100001);
        cout<<a[100000]<<endl;
    }
    return 0;
}

D - Derangement

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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.

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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

5
1 4 3 5 2

Sample Output 1

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.

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Sample Input 2

2
1 2

Sample Output 2

1

Swapping 1 and 2 satisfies the condition.

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Sample Input 3

2
2 1

Sample Output 3

0

The condition is already satisfied initially.

---

Sample Input 4

9
1 2 4 9 5 8 7 3 6

Sample Output 4

3

#include<iostream>
using namespace std;
int a[100010];
int main()
{
	int n,k=0,temp,c=0;
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i];
		if(a[i]!=i)
			k++;
	}
	for(int i=1;i<=n;i++)
	{
		if(a[i]==i)
		{
			temp=a[i];
			a[i]=a[i+1];
			a[i+1]=temp;
			c++;
		}
	}
	if(k==n)
		cout<<0<<endl;
	else
		cout<<c<<endl;
}