CF round#211-2 A (13.11.11)

最新推荐文章于 2022-05-05 16:14:54 发布

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最新推荐文章于 2022-05-05 16:14:54 发布

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本文介绍了一种基于日本传统算盘(Soroban)的数字表示方法，并提供了一个C++程序来展示如何根据输入的数字n，输出该数字在算盘上的具体表示形式。

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You know that Japan is the country with almost the largest 'electronic devices per person' ratio. So you might be quite surprised to find out that the primary school in Japan teaches to count using aSoroban — an abacus developed in Japan. This phenomenon has its reasons, of course, but we are not going to speak about them. Let's have a look at the Soroban's construction.

$\text{[math]}$

Soroban consists of some number of rods, each rod contains five beads. We will assume that the rods are horizontal lines. One bead on each rod (the leftmost one) is divided from the others by a bar (the reckoning bar). This single bead is calledgo-dama and four others are ichi-damas. Each rod is responsible for representing a single digit from 0 to 9. We can obtain the value of a digit by following simple algorithm:

Set the value of a digit equal to 0.
If the go-dama is shifted to the right, add 5.
Add the number of ichi-damas shifted to the left.

Thus, the upper rod on the picture shows digit 0, the middle one shows digit 2 and the lower one shows 7. We will consider the top rod to represent the last decimal digit of a number, so the picture shows number 720.

Write the program that prints the way Soroban shows the given number n.

Input

The first line contains a single integer n (0 ≤ n < 10⁹).

Output

Print the description of the decimal digits of number nfrom the last one to the first one (as mentioned on the picture in the statement), one per line. Print the beads as large English letters 'O', rod pieces as character '-' and the reckoning bar as '|'. Print as many rods, as many digits are in the decimal representation of numbern without leading zeroes. We can assume that number 0 has no leading zeroes.

Sample test(s)

Input

Output

O-|OO-OO

Input

Output

O-|OOO-O
O-|O-OOO

Input

Output

O-|-OOOO
O-|OO-OO
-O|OO-OO

题意：日本的算盘算法，一共五个算珠，算盘分成两块地盘，将五颗算子分成最左边的1个和最右边的4个... 然后最左边的1个要是拨到右边，代表5，然后最右边的4个，往左拨一个算1。现在给出一个数字n，求这个数字n在算盘上是怎么表示的？

思路：就是把各位数都拿来判断一次，先判断是否大于5，然后再判断拨多少个右边的算子...

坑点：n等于0的时候要考虑进去啊～

AC代码：

#include<stdio.h>

int main() {
    int n;
    while(scanf("%d", &n) != EOF) {
        int t;
        if(n == 0)
            printf("O-|-OOOO\n");
        while(n > 0) {
            t = n % 10;
            if(t >= 5) {
                printf("-O|");
                t = t - 5;
                int count = 0;
                while(count++ < 5 ) {
                    if(t-- == 0)
                        printf("-");
                    else
                        printf("O");
                }
                printf("\n");
            }
            else{
                printf("O-|");
                int count = 0;
                while(count++ < 5) {
                    if(t-- == 0)
                        printf("-");
                    else
                        printf("O");
                }
                printf("\n");
            }
            n = n / 10;
        }
    }
    return 0;
}