AtCoder Beginner Contest 100 2018/06/16

A - Happy Birthday!

Time limit : 2sec / Memory limit : 1000MB

Score: 100 points

Problem Statement

E869120's and square1001's 16-th birthday is coming soon.
Takahashi from AtCoder Kingdom gave them a round cake cut into 16 equal fan-shaped pieces.

E869120 and square1001 were just about to eat A and B of those pieces, respectively,
when they found a note attached to the cake saying that "the same person should not take two adjacent pieces of cake".

Can both of them obey the instruction in the note and take desired numbers of pieces of cake?

Constraints

  • A and B are integers between 1 and 16 (inclusive).
  • A+B is at most 16.

Input

Input is given from Standard Input in the following format:

A B

Output

If both E869120 and square1001 can obey the instruction in the note and take desired numbers of pieces of cake, print Yay!; otherwise, print :(.

Sample Input 1

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5 4

Sample Output 1

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Yay!

Both of them can take desired number of pieces as follows: 

Sample Input 2

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8 8

Sample Output 2

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Yay!

Both of them can take desired number of pieces as follows: 

Sample Input 3

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11 4

Sample Output 3

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:(

In this case, there is no way for them to take desired number of pieces, unfortunately.

 代码：

import java.util.*;

public class Main {
    
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int a = in.nextInt();
        int b = in.nextInt();
        if(a <= 8 && b <= 8)System.out.println("Yay!");
        else System.out.println(":(");
    }
}

B - Ringo's Favorite Numbers

Time limit : 2sec / Memory limit : 1000MB

Score: 200 points

Problem Statement

Today, the memorable AtCoder Beginner Contest 100 takes place. On this occasion, Takahashi would like to give an integer to Ringo.
As the name of the contest is AtCoder Beginner Contest 100, Ringo would be happy if he is given a positive integer that can be divided by 100 exactly Dtimes.

Find the N-th smallest integer that would make Ringo happy.

Constraints

  • D is 0, 1 or 2.
  • N is an integer between 1 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

D N

Output

Print the N-th smallest integer that can be divided by 100 exactly D times.

Sample Input 1

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0 5

Sample Output 1

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5

The integers that can be divided by 100 exactly 0 times (that is, not divisible by 100) are as follows: 1, 2, 3, 4, 5, 6, 7, ...
Thus, the 5-th smallest integer that would make Ringo happy is 5.

Sample Input 2

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1 11

Sample Output 2

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1100

The integers that can be divided by 100 exactly once are as follows: 100, 200, 300, 400, 500, 600, 700, 800, 900, 1 000, 1 100, ...
Thus, the integer we are seeking is 1 100.

Sample Input 3

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2 85

Sample Output 3

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850000

The integers that can be divided by 100 exactly twice are as follows: 10 000, 20 000, 30 000, ...
Thus, the integer we are seeking is 850 000.

import java.util.*;

public class Main {
    
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int d = in.nextInt();
        int n = in.nextInt();
        int s = 1;
        for(int i = 0;i < d;i ++) {
            s *= 100;
        }
        if(n == 100)n ++;
        System.out.println(s * n);
    }
}

C - *3 or /2

Time limit : 2sec / Memory limit : 1000MB

Score: 300 points

Problem Statement

As AtCoder Beginner Contest 100 is taking place, the office of AtCoder, Inc. is decorated with a sequence of length N, a={a1,a2,a3,…,aN}.
Snuke, an employee, would like to play with this sequence.

Specifically, he would like to repeat the following operation as many times as possible:

For every i satisfying 1≤i≤N, perform one of the following: "divide ai by 2" and "multiply ai by 3".  
Here, choosing "multiply ai by 3" for every i is not allowed, and the value of ai after the operation must be an integer.

At most how many operations can be performed?

Constraints

  • N is an integer between 1 and 10 000 (inclusive).
  • ai is an integer between 1 and 1 000 000 000 (inclusive).

Input

Input is given from Standard Input in the following format:

N
a1 a2 a3 … aN

Output

Print the maximum number of operations that Snuke can perform.

Sample Input 1

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3
5 2 4

Sample Output 1

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3

The sequence is initially 5,2,4. Three operations can be performed as follows:

  • First, multiply a1 by 3, multiply a2 by 3 and divide a3 by 2. The sequence is now 15,6,2.
  • Next, multiply a1 by 3, divide a2 by 2 and multiply a3 by 3. The sequence is now 45,3,6.
  • Finally, multiply a1 by 3, multiply a2 by 3 and divide a3 by 2. The sequence is now 135,9,3.

Sample Input 2

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4
631 577 243 199

Sample Output 2

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0

No operation can be performed since all the elements are odd. Thus, the answer is 0.

Sample Input 3

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10
2184 2126 1721 1800 1024 2528 3360 1945 1280 1776

Sample Output 3

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39
算一下每个数可以被2整除多少次，总次数。
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 d,m = 0;
        for(int i = 0;i < n;i ++) {
            d = in.nextInt();
            while(d % 2 == 0) {
                d /= 2;
                m ++;
            }
        }
        System.out.println(m);
    }
}

D - Patisserie ABC

Time limit : 2sec / Memory limit : 1000MB

Score: 400 points

Problem Statement

Takahashi became a pastry chef and opened a shop La Confiserie d'ABC to celebrate AtCoder Beginner Contest 100.

The shop sells N kinds of cakes.
Each kind of cake has three parameters "beauty", "tastiness" and "popularity". The i-th kind of cake has the beauty of xi, the tastiness of yi and the popularity of zi.
These values may be zero or negative.

Ringo has decided to have M pieces of cakes here. He will choose the set of cakes as follows:

  • Do not have two or more pieces of the same kind of cake.
  • Under the condition above, choose the set of cakes to maximize (the absolute value of the total beauty) + (the absolute value of the total tastiness) + (the absolute value of the total popularity).

Find the maximum possible value of (the absolute value of the total beauty) + (the absolute value of the total tastiness) + (the absolute value of the total popularity) for the set of cakes that Ringo chooses.

Constraints

  • N is an integer between 1 and 1 000 (inclusive).
  • M is an integer between 0 and N (inclusive).
  • xi,yi,zi (1≤i≤N) are integers between −10 000 000 000 and 10 000 000 000 (inclusive).

Input

Input is given from Standard Input in the following format:

N M
x1 y1 z1
x2 y2 z2
 :  :
xN yN zN

Output

Print the maximum possible value of (the absolute value of the total beauty) + (the absolute value of the total tastiness) + (the absolute value of the total popularity) for the set of cakes that Ringo chooses.

Sample Input 1

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5 3
3 1 4
1 5 9
2 6 5
3 5 8
9 7 9

Sample Output 1

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56

Consider having the 2-nd, 4-th and 5-th kinds of cakes. The total beauty, tastiness and popularity will be as follows:

  • Beauty: 1+3+9=13
  • Tastiness: 5+5+7=17
  • Popularity: 9+8+9=26

The value (the absolute value of the total beauty) + (the absolute value of the total tastiness) + (the absolute value of the total popularity) here is 13+17+26=56. This is the maximum value.

Sample Input 2

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5 3
1 -2 3
-4 5 -6
7 -8 -9
-10 11 -12
13 -14 15

Sample Output 2

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54

Consider having the 1-st, 3-rd and 5-th kinds of cakes. The total beauty, tastiness and popularity will be as follows:

  • Beauty: 1+7+13=21
  • Tastiness: (−2)+(−8)+(−14)=−24
  • Popularity: 3+(−9)+15=9

The value (the absolute value of the total beauty) + (the absolute value of the total tastiness) + (the absolute value of the total popularity) here is 21+24+9=54. This is the maximum value.

Sample Input 3

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10 5
10 -80 21
23 8 38
-94 28 11
-26 -2 18
-69 72 79
-26 -86 -54
-72 -50 59
21 65 -32
40 -94 87
-62 18 82

Sample Output 3

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638

If we have the 3-rd, 4-th, 5-th, 7-th and 10-th kinds of cakes, the total beauty, tastiness and popularity will be −323, 66 and 249, respectively.
The value (the absolute value of the total beauty) + (the absolute value of the total tastiness) + (the absolute value of the total popularity) here is 323+66+249=638. This is the maximum value.

Sample Input 4

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3 2
2000000000 -9000000000 4000000000
7000000000 -5000000000 3000000000
6000000000 -1000000000 8000000000

Sample Output 4

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30000000000

The values of the beauty, tastiness and popularity of the cakes and the value to be printed may not fit into 32-bit integers.

n个蛋糕选m个，要求三项值，每一项的和的绝对值加到一起最大，对于每一项，要么和是正的，要么和是负的取绝对值，一共三项，共八种情况，进行枚举，每个蛋糕八种情况的三项和，排序后，选前m大的，进行更新最大值。

二进制位是1就是和为正，是0就和为负(即把这一项的值当成负的减去变为正的，如果这一项本来就是正的，减去了会更小，肯定不满足，会被值大的情况覆盖掉)。

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
long long va[1000][3];
int main() {
    int n,m,i,j,k;
    scanf("%d%d",&n,&m);
    for (i = 0;i < n;i ++) {
        for (j = 0;j < 3;j ++)
            scanf("%lld",&va[i][j]);
    }
    long long ans = -1;
    for (i = 0;i < 8;i ++) {
        long long sum[1000];///记录每个蛋糕三项和
        for (j = 0;j < n;j ++) {
            sum[j] = 0;///i状态下 sum初始为0
            for (k = 0;k < 3;k ++) {
                if(i >> k & 1)sum[j] += va[j][k];///对应位是1直接加
                else sum[j] -= va[j][k];///否则减，如果结果真的是负的，减去负的就是正的
            }
        }
        sort(sum,sum + n);///默认升序排顺序
        long long tans = 0;
        for (j = 1;j <= m;j ++)
            tans += sum[n - j];///从后往前加m个大的
        ans = max(tans,ans);///更新
    }
    printf("%lld",ans);
    return 0;
}
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原文地址：https://www.cnblogs.com/8023spz/p/9500333.html