POJ2136 Vertical Histogram【打印图案】

Vertical Histogram

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 19827		Accepted: 9483

Description

Write a program to read four lines of upper case (i.e., all CAPITAL LETTERS) text input (no more than 72 characters per line) from the input file and print a vertical histogram that shows how many times each letter (but not blanks, digits, or punctuation) appears in the all-upper-case input. Format your output exactly as shown.

Input

* Lines 1..4: Four lines of upper case text, no more than 72 characters per line.

Output

* Lines 1..??: Several lines with asterisks and spaces followed by one line with the upper-case alphabet separated by spaces. Do not print unneeded blanks at the end of any line. Do not print any leading blank lines. 

Sample Input

THE QUICK BROWN FOX JUMPED OVER THE LAZY DOG.
THIS IS AN EXAMPLE TO TEST FOR YOUR
HISTOGRAM PROGRAM.
HELLO!

Sample Output

                            *
                            *
        *                   *
        *                   *     *   *
        *                   *     *   *
*       *     *             *     *   *
*       *     * *     * *   *     * * *
*       *   * * *     * *   * *   * * * *
*     * * * * * *     * * * * *   * * * *     * *
* * * * * * * * * * * * * * * * * * * * * * * * * *
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

Source

USACO 2003 February Orange

USACO 2003 February Orange

问题链接：POJ2136 Vertical Histogram。

问题简述：参见上述链接。

问题分析：

统计四行输入的大写字母，根据统计结果输出柱状图。

该问题的关键是需要一定的想象力，将统计数据转换成相应的图形。

需要注意的一点是，如果出现次数最多字符的出现次数为max，则输出max行。这是关键的地方。

程序说明：（略）

AC的C++语言程序如下：

/* POJ2136 Vertical Histogram */

#include <iostream>
#include <string>
#include <cstring>
#include <cctype>

using namespace std;

const int MAXN = 26;
int acount[MAXN];

int main()
{
    int linecount, max;
    string s;

    memset(acount, 0, sizeof(acount));

    linecount = 0;
    while (getline(cin, s)) {
        // 统计字母
        for(int i=0; i<(int)s.size(); i++)
//            if(isalpha(s[i]))
//                acount[s[i] - 'A']++;
            if(isupper(s[i]))
                acount[s[i] - 'A']++;

        // 每4行输出一次结果
        if(++linecount == 4) {
            linecount = 0;

            // 计算最大的统计值
            max = 0;
            for(int i=0; i<MAXN; i++)
                if(acount[i] > max)
                    max = acount[i];

            // 输出max行
            for(int i=max; i>0; i--) {
                for(int j=0; j<MAXN; j++) {
                    if(acount[j] >= i)
                        cout << "* ";
                    else
                        cout << "  ";
                }
                cout << endl;
            }

            for(int i=0; i<MAXN; i++)
                cout << (char)('A' + i) << " ";
            cout << endl;
        }
    }

    return 0;
}