AtCoder Beginner Contest 072

今天打得真的很爽，15分钟就AK了，第一次在13名。。。在过生日时打真的是我的幸运日，哈哈哈。

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

Copy

100 17

Sample Output 1

Copy

83

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

有x克，每秒减少一克，问t秒后还有多少克，水题。

 

#include <bits/stdc++.h>
#define ll long long
using namespace std;

int main() {
    int x, t;
    cin >> x >> t;
    printf("%d\n",max(0,x-t));
    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

Copy

atcoder

Sample Output 1

Copy

acdr

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

把位数为奇数的输入下，水题。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e5+10;
char str[N];
int main() {
    cin >> str;
    for(int i = 0; str[i];  i++) {
        if(i %2 == 0) printf("%c",str[i]);
    }
    printf("\n");
    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,…,a_N.

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

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

Maximize this count by making optimal choices.

Constraints

• 1≤N≤10⁵
• 0≤a_i<10⁵(1≤i≤N)
• a_i 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 a_i=X.

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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.

 

每位可以加减一或不动，求相同数字的个数最大值。水题。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e5+10;
// int a[N];
map<int, int> mp;
int main() {
    int n, x;
    scanf("%d",&n);
    for(int i = 0; i < n; i ++) {
        scanf("%d",&x);
        mp[x]++;
        mp[x+1]++;
        mp[x-1]++;
    }
    map<int, int> ::iterator it;
    int MAX = 1;
    for(it = mp.begin(); it != mp.end(); ++it) {
        if(MAX < (*it).second){
            MAX = (*it).second;
        }
    }
    printf("%d\n",MAX);
    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,…,p_N 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 p_i≠i for all 1≤i≤N. Find the minimum required number of operations to achieve this.

Constraints

• 2≤N≤10⁵
• p₁,p₂,..,p_N 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

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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.

 

每次可以和相邻的交换位置，求最少的交换次数使的a[i] != i,继续水题。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e5+10;
int a[N];
int main() {
    int n;
    scanf("%d",&n);
    for(int i = 1; i <= n; i ++) {
        scanf("%d", &a[i]);
    }
    int ans = 0;
    for(int i = 1; i < n; i ++) {
        if(a[i] == i) {
            swap(a[i], a[i+1]);
            ans++;
        }
    }
    if(a[n] == n) ans++;
    printf("%d\n",ans);
    return 0;
}