AtCoder Beginner Contest 077(ABC)

A - Rotation

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Time limit : 2sec / Memory limit : 256MB

Score : 100 points

Problem Statement

You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_ij.

Write a program that prints YES if this grid remains the same when rotated 180 degrees, and prints NO otherwise.

Constraints

• C_i,j(1≤i≤2,1≤j≤3) is a lowercase English letter.

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Input

Input is given from Standard Input in the following format:

C11C12C13
C21C22C23

Output

Print YES if this grid remains the same when rotated 180 degrees; print NO otherwise.

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Sample Input 1

pot
top

Sample Output 1

YES

This grid remains the same when rotated 180 degrees.

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Sample Input 2

tab
bet

Sample Output 2

NO

This grid does not remain the same when rotated 180 degrees.

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Sample Input 3

eye
eel

Sample Output 3

NO

    #include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <algorithm>
     
    using namespace std;
    string s,s1;
    int main(){
        cin>>s>>s1;
        bool prime=true;
        int slen=s.length();
        int sslen=s1.length();
        for(int i=0,j=sslen-1;i<slen&&j>=0;i++,j--){
            if(s[i]!=s1[j])
                {
                    prime=false;
                    break;
                }
        }
        if(prime)
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
        return 0;
    }

B - Around Square

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Time limit : 2sec / Memory limit : 256MB

Score : 200 points

Problem Statement

Find the largest square number not exceeding N. Here, a square number is an integer that can be represented as the square of an integer.

Constraints

• 1≤N≤10⁹
• N is an integer.

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Input

Input is given from Standard Input in the following format:

N

Output

Print the largest square number not exceeding N.

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Sample Input 1

10

Sample Output 1

9

10 is not square, but 9=3×3 is. Thus, we print 9.

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Sample Input 2

81

Sample Output 2

81

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Sample Input 3

271828182

Sample Output 3

271821169

    #include <iostream>
    #include <cmath>
    #include <cstring>
    #include <cstdio>
     
    using namespace std;
    int n;
    int main(){
        while(cin>>n){
            int a=sqrt(n);
            cout<<a*a<<endl;
        }
        return 0;
    }

C - Snuke Festival

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Time limit : 2sec / Memory limit : 256MB

Score : 300 points

Problem Statement

The season for Snuke Festival has come again this year. First of all, Ringo will perform a ritual to summon Snuke. For the ritual, he needs an altar, which consists of three parts, one in each of the three categories: upper, middle and lower.

He has N parts for each of the three categories. The size of the i-th upper part is A_i, the size of the i-th middle part is B_i, and the size of the i-th lower part is C_i.

To build an altar, the size of the middle part must be strictly greater than that of the upper part, and the size of the lower part must be strictly greater than that of the middle part. On the other hand, any three parts that satisfy these conditions can be combined to form an altar.

How many different altars can Ringo build? Here, two altars are considered different when at least one of the three parts used is different.

Constraints

• 1≤N≤10⁵
• 1≤A_i≤10⁹(1≤i≤N)
• 1≤B_i≤10⁹(1≤i≤N)
• 1≤C_i≤10⁹(1≤i≤N)
• All input values are integers.

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Input

Input is given from Standard Input in the following format:

N
A1 … AN
B1 … BN
C1 … CN

Output

Print the number of different altars that Ringo can build.

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Sample Input 1

2
1 5
2 4
3 6

Sample Output 1

3

The following three altars can be built:

• Upper: 1-st part, Middle: 1-st part, Lower: 1-st part
• Upper: 1-st part, Middle: 1-st part, Lower: 2-nd part
• Upper: 1-st part, Middle: 2-nd part, Lower: 2-nd part

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Sample Input 2

3
1 1 1
2 2 2
3 3 3

Sample Output 2

27

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Sample Input 3

6
3 14 159 2 6 53
58 9 79 323 84 6
2643 383 2 79 50 288

Sample Output 3

87

    #include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <algorithm>
    #include <vector>
    #define ll long long int
    #define N 100005
    using namespace std;
    vector<int> k[3];
    ll dp[3][N];
    int main(){
        int n;
        cin>>n;
        for(int i=0;i<3;i++){
            for(int j=0;j<n;j++){
                int x;
                cin>>x;
                k[i].push_back(x);
            }
            sort(k[i].begin(),k[i].end());
        }
        dp[0][0]=1;
        for(int i=1;i<n;i++)
            dp[0][i]=dp[0][i-1]+1;
     
        for(int i=1;i<3;i++){
            for(int j=0;j<n;j++){
                int x=lower_bound(k[i-1].begin(),k[i-1].end(),k[i][j])-k[i-1].begin();
                dp[i][j]+=dp[i-1][x-1];
            }
            for(int t=1;t<n;t++)
                dp[i][t]+=dp[i][t-1];
        }
        cout<<dp[2][n-1]<<endl;
        return 0;
    }