• Codeforces Round #135 (Div. 2)---A. k-String


    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    A string is called a k-string if it can be represented as k concatenated copies of some string. For example, the string "aabaabaabaab" is at the same time a 1-string, a 2-string and a 4-string, but it is not a 3-string, a 5-string, or a 6-string and so on. Obviously any string is a 1-string.

    You are given a string s, consisting of lowercase English letters and a positive integer k. Your task is to reorder the letters in the string sin such a way that the resulting string is a k-string.

    Input

    The first input line contains integer k (1 ≤ k ≤ 1000). The second line contains s, all characters in s are lowercase English letters. The string length s satisfies the inequality 1 ≤ |s| ≤ 1000, where |s| is the length of string s.

    Output

    Rearrange the letters in string s in such a way that the result is a k-string. Print the result on a single output line. If there are multiple solutions, print any of them.

    If the solution doesn't exist, print "-1" (without quotes).

    Sample test(s)
    input
    2
    aazz
    
    output
    azaz
    
    input
    3
    abcabcabz
    
    output
    -1





    解题思路:给一个串。问能否由k个同样的串连接而成。

    用STL里的map。扫一遍。分别记录每一个字符的个数。在推断全部的字符是否是k的倍数,若不是,则输出-1;否则,遍历依次map。每一个字符输出(总个数)/k个,然后反复k次就可以。





    AC代码:

    #include <stdio.h>
    #include <string.h>
    #include <iostream>
    #include <algorithm>
    #include <vector>
    #include <queue>
    #include <set>
    #include <map>
    #include <string>
    #include <math.h>
    #include <stdlib.h>
    #include <time.h>
    using namespace std;
    #define INF 0x7fffffff
    
    map<char, int> m;
    
    int main()
    {
        #ifdef sxk
            freopen("in.txt","r",stdin);
        #endif
        int n;
        string s;
        while(scanf("%d",&n)!=EOF)
        {
            cin>>s;
            int len = s.size();
            for(int i=0; i<len; i++)
                m[s[i]] ++;
           map<char, int>::iterator it;
           int flag = 1;
            for(it=m.begin(); it!=m.end(); it++){
                if(it->second % n){
                    flag = 0;
                    break;
                }
            }
            if(!flag)  printf("-1
    ");
            else{
                for(int j=0; j<n; j++){
                    for(it=m.begin(); it!=m.end(); it++){
                        for(int i=1; i<=it->second/n; i++)
                            printf("%c", it->first);
                    }
                }
                printf("
    ");
            }
        }
        return 0;
    }


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  • 原文地址:https://www.cnblogs.com/mthoutai/p/6867404.html
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