zoukankan      html  css  js  c++  java
  • Codeforces 263B. Appleman and Card Game

    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 ith, 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
    82
    input
    6 4
    YJSNPI
    output
    4
     
    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.

    解题:贪心。

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <cmath>
     5 #include <algorithm>
     6 #include <climits>
     7 #include <vector>
     8 #include <queue>
     9 #include <cstdlib>
    10 #include <string>
    11 #include <set>
    12 #include <stack>
    13 #define LL long long
    14 #define pii pair<int,int>
    15 #define INF 0x3f3f3f3f
    16 using namespace std;
    17 int letter[27];
    18 char str[100100];
    19 bool cmp(const int &x,const int &y){
    20     return x > y;
    21 }
    22 int main() {
    23     int n,k;
    24     LL ans;
    25     while(~scanf("%d %d",&n,&k)){
    26         memset(letter,0,sizeof(letter));
    27         scanf("%s",str);
    28         for(int i = 0; str[i]; i++) letter[str[i]-'A']++;
    29         sort(letter,letter+26,cmp);
    30         for(int i = ans = 0; k && i < 26; i++){
    31             ans += 1LL*min(k,letter[i])*min(k,letter[i]);
    32             k -= min(k,letter[i]);
    33         }
    34         cout<<ans<<endl;
    35     }
    36     return 0;
    37 }
    View Code
  • 相关阅读:
    libstdc++.so.5: undefined reference to `memcpy@GLIBC_2.14'
    Redis开启远程访问
    Centos7.5 防火墙设置
    Redis和MemCache的区别
    Redis的事务
    Redis持久化——AOF
    Redis持久化——RDB快照
    Redis配置文件介绍
    Redis常见操作命令
    Redis的五大数据类型
  • 原文地址:https://www.cnblogs.com/crackpotisback/p/3938682.html
Copyright © 2011-2022 走看看