CF#263

冷不防 2022-03-31 10:38 157阅读 0赞

昨天没打，今天写了一下，前三题都没有难度吧。

A. Appleman and Easy Task

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?

Given a *n* × *n* checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the board are adjacent if they share a side.

Input

The first line contains an integer *n* (1 ≤ *n* ≤ 100). Then *n* lines follow containing the description of the checkerboard. Each of them contains *n* characters (either 'x' or 'o') without spaces.

Output

Print "YES" or "NO" (without the quotes) depending on the answer to the problem.

Sample test(s)

input

3
    xxo
    xox
    oxx

output

YES

input

4
    xxxo
    xoxo
    oxox
    xxxx

output

题意：推断一个点的上下左右四个方向’o'的个数，都是偶数输出YES。否则输出NO。

#include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<limits.h>
    #include<vector>
    typedef long long LL;
    using namespace std;
    char mp[110][110];
    int n;
    
    int main()
    {
        while(cin>>n)
        {
            int ok=1;
            for(int i=1;i<=n;i++)
            {
                for(int j=1;j<=n;j++)
                    cin>>mp[i][j];
            }
            for(int i=1;i<=n;i++)
            {
                for(int j=1;j<=n;j++)
                {
                    int cnt=0;
                    if(mp[i+1][j]=='o')
                        cnt++;
                    if(mp[i-1][j]=='o')
                        cnt++;
                    if(mp[i][j-1]=='o')
                        cnt++;
                    if(mp[i][j+1]=='o')
                        cnt++;
                    if(cnt&1)
                    {
                        ok=0;
                        break;
                    }
                }
            }
            if(ok)  cout<<"YES"<<endl;
            else   cout<<"NO"<<endl;
        }
        return 0;
    }

B. Appleman and Card Game

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Appleman has *n* cards. Each card has an uppercase letter written on it. Toastman must choose *k* cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card *i* you should calculate how much Toastman's cards have the letter equal to letter on *i*th, then sum up all these quantities, such a number of coins Appleman should give to Toastman.

Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?

Input

The first line contains two integers *n* and *k* (1 ≤ *k* ≤ *n* ≤ 105). The next line contains *n* uppercase letters without spaces — the *i*\-th letter describes the *i*\-th card of the Appleman.

Output

Print a single integer – the answer to the problem.

Sample test(s)

input

15 10
    DZFDFZDFDDDDDDF

output

input

6 4
    YJSNPI

output

Note

In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin.

水题，hash后排序计算就可以。

#include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<limits.h>
    #include<vector>
    typedef long long LL;
    using namespace std;
    int hash[30];
    int cmp(int a,int b)
    {
        return a>b;
    }
    int main()
    {
        int n,k;
        char  c;
        while(cin>>n>>k)
        {
            memset(hash,0,sizeof(hash));
            for(int i=0;i<n;i++)
            {
                cin>>c;
                hash[c-'A']++;
            }
            sort(hash,hash+26,cmp);
            LL sum=0;
            for(int i=0;i<26;i++)
            {
                if(k==0)   break;
                if(k>=hash[i])
                {
                    sum+=(LL)hash[i]*hash[i];
                    k-=hash[i];
                }
                else
                {
                    sum+=(LL)k*k;
                    k=0;
                }
            }
            cout<<sum<<endl;
        }
        return 0;
    }

C. Appleman and Toastman

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Appleman and Toastman play a game. Initially Appleman gives one group of *n* numbers to the Toastman, then they start to complete the following tasks:

*  Each time Toastman gets a group of numbers, he sums up all the numbers and adds this sum to the score. Then he gives the group to the Appleman.
 *  Each time Appleman gets a group consisting of a single number, he throws this group out. Each time Appleman gets a group consisting of more than one number, he splits the group into two non-empty groups (he can do it in any way) and gives each of them to Toastman.

After guys complete all the tasks they look at the score value. What is the maximum possible value of score they can get?

Input

The first line contains a single integer *n* (1 ≤ *n* ≤ 3·105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1 ≤ *a**i* ≤ 106) — the initial group that is given to Toastman.

Output

Print a single integer — the largest possible score.

Sample test(s)

input

3
    3 1 5

output

input

1
    10

output

Note

Consider the following situation in the first example. Initially Toastman gets group \[3, 1, 5\] and adds 9 to the score, then he give the group to Appleman. Appleman splits group \[3, 1, 5\] into two groups: \[3, 5\] and \[1\]. Both of them should be given to Toastman. When Toastman receives group \[1\], he adds 1 to score and gives the group to Appleman (he will throw it out). When Toastman receives group \[3, 5\], he adds 8 to the score and gives the group to Appleman. Appleman splits \[3, 5\] in the only possible way: \[5\] and \[3\]. Then he gives both groups to Toastman. When Toastman receives \[5\], he adds 5 to the score and gives the group to Appleman (he will throws it out). When Toastman receives \[3\], he adds 3 to the score and gives the group to Appleman (he will throws it out). Finally Toastman have added 9 + 1 + 8 + 5 + 3 = 26 to the score. This is the optimal sequence of actions.

计数问题，每次抛出一个最小的数就可以得到最大值。

#include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<limits.h>
    #include<vector>
    typedef long long LL;
    using namespace std;
    LL num[300000+100];
    int main()
    {
        int n;
        while(~scanf("%d",&n))
        {
            LL sum=0;
            for(int i=1;i<=n;i++)
                scanf("%d",&num[i]);
            sort(num+1,num+n+1);
            for(int i=1;i<=n;i++)
                sum+=num[i]*(i+1);
            cout<<sum-num[n]<<endl;
        }
        return 0;
    }