AtCoder Beginner Contest 071 ABCD

A - Meal Delivery

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Time limit : 2sec / Memory limit : 256MB

Score : 100 points

Problem Statement

Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery.

Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence.

Here, the distance between two points s and t on a number line is represented by |s−t|.

Constraints

• 1≤x≤1000
• 1≤a≤1000
• 1≤b≤1000
• x,a and b are pairwise distinct.
• The distances between Snuke's residence and stores A and B are different.

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Input

Input is given from Standard Input in the following format:

x a b

Output

If store A is closer, print A; if store B is closer, print B.

---

Sample Input 1

Copy

5 2 7

Sample Output 1

Copy

B

The distances between Snuke's residence and stores A and B are 3 and 2, respectively. Since store B is closer, print B.

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Sample Input 2

Copy

1 999 1000

Sample Output 2

Copy

A

求x到a或b的最近的那个，水题。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 110;
int main() {
    int x, a, b;
    cin >> x >> a >> b;
    if(abs(x-a) < abs(x-b)) printf("A
");
    else printf("B
");
    return 0;
}

B - Not Found

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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. Find the lexicographically (alphabetically) smallest lowercase English letter that does not occur in S. If every lowercase English letter occurs in S, print None instead.

Constraints

• 1≤|S|≤10⁵ (|S| is the length of string S.)
• S consists of lowercase English letters.

---

Input

Input is given from Standard Input in the following format:

S

Output

Print the lexicographically smallest lowercase English letter that does not occur in S. If every lowercase English letter occurs in S, print None instead.

---

Sample Input 1

Copy

atcoderregularcontest

Sample Output 1

Copy

b

The string atcoderregularcontest contains a, but does not contain b.

---

Sample Input 2

Copy

abcdefghijklmnopqrstuvwxyz

Sample Output 2

Copy

None

This string contains every lowercase English letter.

---

Sample Input 3

Copy

fajsonlslfepbjtsaayxbymeskptcumtwrmkkinjxnnucagfrg

Sample Output 3

Copy

d

从a到z，看哪个没出现就输出谁，都出现了就输出None，接着水。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e5+10;
char str[N];
int a[30];
int main() {
    cin >> str;
    for(int i = 0; str[i]; i ++) a[str[i]-'a']++;
    for(int i = 0; i < 26; i ++) {
        if(a[i] == 0) return 0*printf("%c
",i+'a');
    }
    printf("None
");
    return 0;
}

C - Make a Rectangle

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Time limit : 2sec / Memory limit : 256MB

Score : 300 points

Problem Statement

We have N sticks with negligible thickness. The length of the i-th stick is A_i.

Snuke wants to select four different sticks from these sticks and form a rectangle (including a square), using the sticks as its sides. Find the maximum possible area of the rectangle.

Constraints

• 4≤N≤10⁵
• 1≤A_i≤10⁹
• A_i is an integer.

---

Input

Input is given from Standard Input in the following format:

N
A1 A2 ... AN

Output

Print the maximum possible area of the rectangle. If no rectangle can be formed, print 0.

---

Sample Input 1

Copy

6
3 1 2 4 2 1

Sample Output 1

Copy

2

1×2 rectangle can be formed.

---

Sample Input 2

Copy

4
1 2 3 4

Sample Output 2

Copy

0

No rectangle can be formed.

---

Sample Input 3

Copy

10
3 3 3 3 4 4 4 5 5 5

Sample Output 3

Copy

20

求最大的两个出现两次出现过的数字，继续水。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 110;
map<int,int> mp;
int main() {
    int n, x;

    cin >> n;
    int MAX1 = 0, MAX2 = 0;
    for(int i = 0; i < n; i ++) {
        scanf("%d", &x);
        if(mp.count(x)) {
            if(x > MAX1) {
                MAX2 = MAX1;
                MAX1 = x;
            } else if(x > MAX2) {
                if(x == MAX1 && mp[x] >= 3) {
                    MAX2 = x;
                } else if(x < MAX1) MAX2 = x;
            }
        }
        mp[x] ++;
    }
    printf("%lld
",1LL*MAX1*MAX2);
    return 0;
}

D - Coloring Dominoes

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Time limit : 2sec / Memory limit : 256MB

Score : 400 points

Problem Statement

We have a board with a 2×N grid. Snuke covered the board with N dominoes without overlaps. Here, a domino can cover a 1×2 or 2×1 square.

Then, Snuke decided to paint these dominoes using three colors: red, cyan and green. Two dominoes that are adjacent by side should be painted by different colors. Here, it is not always necessary to use all three colors.

Find the number of such ways to paint the dominoes, modulo 1000000007.

The arrangement of the dominoes is given to you as two strings S₁ and S₂ in the following manner:

• Each domino is represented by a different English letter (lowercase or uppercase).
• The j-th character in S_i represents the domino that occupies the square at the i-th row from the top and j-th column from the left.

Constraints

• 1≤N≤52
• |S₁|=|S₂|=N
• S₁ and S₂ consist of lowercase and uppercase English letters.
• S₁ and S₂ represent a valid arrangement of dominoes.

---

Input

Input is given from Standard Input in the following format:

N
S1
S2

Output

Print the number of such ways to paint the dominoes, modulo 1000000007.

---

Sample Input 1

Copy

3
aab
ccb

Sample Output 1

Copy

6

There are six ways as shown below:

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Sample Input 2

Copy

1
Z
Z

Sample Output 2

Copy

3

Note that it is not always necessary to use all the colors.

---

Sample Input 3

Copy

52
RvvttdWIyyPPQFFZZssffEEkkaSSDKqcibbeYrhAljCCGGJppHHn
RLLwwdWIxxNNQUUXXVVMMooBBaggDKqcimmeYrhAljOOTTJuuzzn

Sample Output 3

Copy

958681902

这题判断下就行了。
四种情况。
1、
aac
bbc
当假设判断到c了,ans*=1, a为1，b为2时，c只能为3
2、
aabb
ccdd
当判断到bb这时,ans*=3 假设a为1，c为2时  那个b和d有三种情况 （2,1）（2,3）（3,1）
3、
abb
acc
当判断到bb时,ans*=2 a为1时 b和c有（2,3）（3,2）
4、
ab
ab
当判断到b时，ans*=2,a为1时，b有 2,3


但在首个位置特殊判断下就行。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const ll N = 110;
const ll mod = 1000000007;
char str[N], str1[N];
int main() {
    ll n, ans = 1;
    cin >> n >> str >> str1;
    bool flag = false;
    for(ll i = 0; i < n; i ++) {
        if(i == 0) {
            if(str[i] == str1[i]) {
                ans = 3;
            } else {
                ans = 6;flag = true;i++;
            }
        }else {
            if(str[i] == str1[i]) {
                if(flag) {
                    flag = false;
                }else {
                    ans *= 2;
                    ans %= mod;
                }
                flag = false;
            } else {
                if(flag) {
                    ans *= 3;
                    ans %= mod;
                }else {
                    ans *= 2;
                    ans %= mod;
                }
                flag = true;
                i++;
            }
        }
    }
    printf("%lld
",ans%mod);
    return 0;
}