2016亚洲区域赛现场赛北京赛区f题

最新推荐文章于 2019-11-02 11:53:59 发布

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本文介绍了一个简单的音乐统计问题，通过分析由传统乐器演奏的中国古代音乐片段，统计出现频率最高的五声音阶（宫、商、角、徵、羽）。使用C语言实现了一种统计方法，并按出现次数从高到低输出，若次数相同则按ASCII码顺序输出。

Problem F. What a Simple Research
Description
Peking University Student Folk Music Band has a history of more than 90 years. They play Chinese traditional music by Chinese traditional instruments, such as Pipa, Erhu and Guzheng, etc. Doctor Li is a member of that band, and also a former ACMer. Now he is doing some research on Chinese ancient music. Many Chinese ancient music has only five kinds of tones, which can be denoted by ‘C’,’D’,’E’,’G’, and ‘A’. Given a piece of music score, Li wants to do some simple statistics.
Input
There are no more than 20 test cases.
In each test case:
The first line contains two integers n and m (2<= n,m <= 20), indicating that a piece of music score is represented by an n×m matrix of tones. Only ‘C’,’D’,’E’,’G’ and ‘A’ can appear in the matrix.
Then the n×m matrix follows.
The input ends with a line of “0 0”.
Output
For each test case:
For each kind of tone shown in the matrix, calculate the appearing times of it, and print the result in descending order according to the appearing times. If more than one kind of tones has the same appearing times, print them in the lexicographical order.
Sample Input
4 5
AGCDE
AGDDE
DDDDD
EEEEE
2 4
GADC
CDEE
0 0
Sample Output
D 8 E 7 A 2 G 2 C 1
C 2 D 2 E 2 A 1 G 1
题意：给你一个字符矩阵，里面只含有A,C,D,E,G五种字符，问你把出现过的字符次数从大到小排出来，若次数相同，则按照acs码从大到小。

解题思路：签到题，模拟即可，注意没有出现的字符不要输出来就行

#include<stdio.h>
#include<algorithm>
using namespace std;
struct node{
    int value;
    char ch;
}Node[10];
bool cmp(node n1,node n2)
{
    if(n1.value == n2.value) return n1.ch < n2.ch;
    else return n1.value > n2.value;
}
int n,m;
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        if(n == 0&&m == 0) break;
        char x;
        Node[1].value = 0;
        Node[1].ch = 'A';
        Node[2].value = 0;
        Node[2].ch = 'C';
        Node[3].value = 0;
        Node[3].ch = 'D';
        Node[4].value = 0;
        Node[4].ch = 'E';
        Node[5].value = 0;
        Node[5].ch = 'G';
        getchar();
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= m; j++)
            {
                scanf("%c",&x);
                for(int k = 1; k <= 5; k++)
                {
                    if(x == Node[k].ch)
                    {
                        Node[k].value++;
                        break;
                    }
                }
            }
            getchar();
        }
        sort(Node + 1,Node + 6,cmp);
        bool flag = true;
        for(int i = 1; i <= 5; i++)
        {
            if(Node[i].value != 0)
            {
                if(flag)
                {
                    printf("%c %d",Node[i].ch,Node[i].value);
                    flag = false;
                }
                else printf(" %c %d",Node[i].ch,Node[i].value);
            }
        }
        printf("\n");
    }
    return 0;
}