AtCoder Beginner Contest 102

A - Multiple of 2 and N

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Time Limit: 2 sec / Memory Limit: 1024 MB

Score : $100100$ points

Problem Statement

You are given a positive integer $\mathrm{NN}$. Find the minimum positive integer divisible by both $22$ and $\mathrm{NN}$.

Constraints

• $\mathrm{1\le N\le 1091\le N\le 109}$
• All values in input are integers.

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Input

Input is given from Standard Input in the following format:

$\mathrm{NN}$

Output

Print the minimum positive integer divisible by both $22$ and $\mathrm{NN}$.

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Sample Input 1 Copy

Copy

3

Sample Output 1 Copy

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6

$66$ is divisible by both $22$ and $33$. Also, there is no positive integer less than $66$ that is divisible by both $22$ and $33$. Thus, the answer is $66$.

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Sample Input 2 Copy

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10

Sample Output 2 Copy

Copy

10

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Sample Input 3 Copy

Copy

999999999

Sample Output 3 Copy

Copy

1999999998
代码：

import java.util.*;

public class Main {
    
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        if(n % 2 == 1)n *= 2;
        System.out.println(n);
    }
}

B - Maximum Difference

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Time Limit: 2 sec / Memory Limit: 1024 MB

Score : $200200$ points

Problem Statement

You are given an integer sequence $\mathrm{AA}$ of length $\mathrm{NN}$. Find the maximum absolute difference of two elements (with different indices) in $\mathrm{AA}$.

Constraints

• $\mathrm{2\le N\le 1002\le N\le 100}$
• $\mathrm{1\le Ai\le 1091\le Ai\le 109}$
• All values in input are integers.

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Input

Input is given from Standard Input in the following format:

$\mathrm{NN}$
${\mathrm{A1A1}}^{}$ ${\mathrm{A2A2}}^{}$ $......$ ${\mathrm{ANAN}}^{}$

Output

Print the maximum absolute difference of two elements (with different indices) in $\mathrm{AA}$.

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Sample Input 1 Copy

Copy

4
1 4 6 3

Sample Output 1 Copy

Copy

5

The maximum absolute difference of two elements is ${\mathrm{A3-A1=6-1=5A3-A1=6-1=5}}^{}$.

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Sample Input 2 Copy

Copy

2
1000000000 1

Sample Output 2 Copy

Copy

999999999

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Sample Input 3 Copy

Copy

5
1 1 1 1 1

Sample Output 3 Copy

Copy

0
代码：

import java.util.*;

public class Main {
    
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int max = 0,min = 1000000000;
        for(int i = 0;i < n;i ++) {
            int d = in.nextInt();
            max = Math.max(max, d);
            min = Math.min(min, d);
        }
        System.out.println(max - min);
    }
}

C - Linear Approximation

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Time Limit: 2 sec / Memory Limit: 1024 MB

Score : $300300$ points

Problem Statement

Snuke has an integer sequence $\mathrm{AA}$ of length $\mathrm{NN}$.

He will freely choose an integer $\mathrm{bb}$. Here, he will get sad if ${\mathrm{AiAi}}^{}$ and $\mathrm{b+ib+i}$ are far from each other. More specifically, the sadness of Snuke is calculated as follows:

• $\mathrm{abs(A1-(b+1))+abs(A2-(b+2))+...+abs(AN-(b+N))abs(A1-(b+1))+abs(A2-(b+2))+...+abs(AN-(b+N))}$

Here, $\mathrm{abs(x)abs(x)}$ is a function that returns the absolute value of $\mathrm{xx}$.

Find the minimum possible sadness of Snuke.

Constraints

• $\mathrm{1\le N\le 2\times 1051\le N\le 2\times 105}$
• $\mathrm{1\le Ai\le 1091\le Ai\le 109}$
• All values in input are integers.

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Input

Input is given from Standard Input in the following format:

$\mathrm{NN}$
${\mathrm{A1A1}}^{}$ ${\mathrm{A2A2}}^{}$ $......$ ${\mathrm{ANAN}}^{}$

Output

Print the minimum possible sadness of Snuke.

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Sample Input 1 Copy

Copy

5
2 2 3 5 5

Sample Output 1 Copy

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2

If we choose $\mathrm{b=0b=0}$, the sadness of Snuke would be $\mathrm{abs(2-(0+1))+abs(2-(0+2))+abs(3-(0+3))+abs(5-(0+4))+abs(5-(0+5))=2abs(2-(0+1))+abs(2-(0+2))+abs(3-(0+3))+abs(5-(0+4))+abs(5-(0+5))=2}$. Any choice of $\mathrm{bb}$ does not make the sadness of Snuke less than $22$, so the answer is $22$.

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Sample Input 2 Copy

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9
1 2 3 4 5 6 7 8 9

Sample Output 2 Copy

Copy

0

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Sample Input 3 Copy

Copy

6
6 5 4 3 2 1

Sample Output 3 Copy

Copy

18

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Sample Input 4 Copy

Copy

7
1 1 1 1 2 3 4

Sample Output 4 Copy

Copy

6
先把所有的值减去对应下标的值，之后的序列是求都减去同一个值后，绝对值的和最小，那么就排序，找到中间位置的值，所有的数减去中间位置的值就可以了，中间位置的左边比他小，右边比他大，减去值后中间的变为0，左边的都增加，右边的豆腐减少，保证减少的小于等于增加的即可。
代码：

import java.util.*;

public class Main {
    
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int [] s = new int[n];
        long d = 0;
        for(int i = 0;i < n;i ++) {
            s[i] = in.nextInt();
            s[i] -= i + 1;
        }
        Arrays.sort(s);
        for(int i = 0;i < n;i ++) {
            d += Math.abs(s[i] - s[n / 2]);
        }
        System.out.println(d);
    }
}

D - Equal Cut

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Time Limit: 2 sec / Memory Limit: 1024 MB

Score : $600600$ points

Problem Statement

Snuke has an integer sequence $\mathrm{AA}$ of length $\mathrm{NN}$.

He will make three cuts in $\mathrm{AA}$ and divide it into four (non-empty) contiguous subsequences $\mathrm{B,C,DB,C,D}$ and $\mathrm{EE}$. The positions of the cuts can be freely chosen.

Let $\mathrm{P,Q,R,SP,Q,R,S}$ be the sums of the elements in $\mathrm{B,C,D,EB,C,D,E}$, respectively. Snuke is happier when the absolute difference of the maximum and the minimum among $\mathrm{P,Q,R,SP,Q,R,S}$ is smaller. Find the minimum possible absolute difference of the maximum and the minimum among $\mathrm{P,Q,R,SP,Q,R,S}$.

Constraints

• $\mathrm{4\le N\le 2\times 1054\le N\le 2\times 105}$
• $\mathrm{1\le Ai\le 1091\le Ai\le 109}$
• All values in input are integers.

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Input

Input is given from Standard Input in the following format:

$\mathrm{NN}$
${\mathrm{A1A1}}^{}$ ${\mathrm{A2A2}}^{}$ $......$ ${\mathrm{ANAN}}^{}$

Output

Find the minimum possible absolute difference of the maximum and the minimum among $\mathrm{P,Q,R,SP,Q,R,S}$.

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Sample Input 1 Copy

Copy

5
3 2 4 1 2

Sample Output 1 Copy

Copy

2

If we divide $\mathrm{AA}$ as $\mathrm{B,C,D,E=(3),(2),(4),(1,2)B,C,D,E=(3),(2),(4),(1,2)}$, then $\mathrm{P=3,Q=2,R=4,S=1+2=3P=3,Q=2,R=4,S=1+2=3}$. Here, the maximum and the minimum among $\mathrm{P,Q,R,SP,Q,R,S}$are $44$ and $22$, with the absolute difference of $22$. We cannot make the absolute difference of the maximum and the minimum less than $22$, so the answer is $22$.

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Sample Input 2 Copy

Copy

10
10 71 84 33 6 47 23 25 52 64

Sample Output 2 Copy

Copy

36

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Sample Input 3 Copy

Copy

7
1 2 3 1000000000 4 5 6

Sample Output 3 Copy

Copy

999999994
如果是排着去找这三个位置，是O(n^3)的，可以采用折半枚举的技巧，先找到第二个位置，然后两边各自分为两块，用l和r标记，两块的差距尽可能小即可，随着i的右移，lr只需要初始化一次，然后跟着移动即可，i的移动必然会打破原来的平衡，lr只能继续右移，不需要从最初端开始，不然仍然超时。
java代码：

import java.util.*;

public class Main {
    
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        long [] s = new long[n + 1];
        for(int i = 1;i <= n;i ++) {
            s[i] = in.nextLong();
            s[i] += s[i - 1];
        }
        int l = 1,r = 3;
        long ans = s[n];
        for(int i = 2;i < n - 1;i ++) {
            long last = s[n];
            while(l < i) {
                long d = Math.abs(s[i] - s[l] * 2);
                if(d > last) break;
                else last = d;
                l ++;
            }
            l --;
            last = s[n];
            while(r < n) {
                long d = Math.abs(s[n] + s[i] - s[r] * 2);
                if(d > last) break;
                else last = d;
                r ++;
            }
            r --;
            long max = Math.max(Math.max(Math.max(s[l], s[i] - s[l]), s[r] - s[i]), s[n] - s[r]);
            long min = Math.min(Math.min(Math.min(s[l], s[i] - s[l]), s[r] - s[i]), s[n] - s[r]);
            ans = Math.min(max - min, ans);
        }
        System.out.println(ans);
    }
}

c++代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define Max 200005
using namespace std;
typedef long long LL;
int main() {
    int n;
    int a[Max];
    LL sum[Max] = {0};
    scanf("%d",&n);
    for(int i = 1;i <= n;i ++) {
        scanf("%d",&a[i]);
        sum[i] = sum[i - 1] + a[i];
    }
    LL ans = sum[n];
    int l = 1,r = 3;
    for(int i = 2;i <= n - 2;i ++) {
        LL last = sum[n];
        while(l < i) {
            LL d = abs(sum[i] - sum[l] * 2);
            if(d > last) break;
            else last = d;
            l ++;
        }
        l --;
        last = sum[n];
        while(r < n) {
            LL d = abs(sum[n] + sum[i] - sum[r] * 2);
            if(d > last) break;
            else last = d;
            r ++;
        }
        r --;
        LL up = max(max(max(sum[l],sum[i] - sum[l]),sum[r] - sum[i]),sum[n] - sum[r]);
        LL down = min(min(min(sum[l],sum[i] - sum[l]),sum[r] - sum[i]),sum[n] - sum[r]);
        ans = min(ans,up - down);
    }
    printf("%lld",ans);
}