1005. Spell It Right (20)

最新推荐文章于 2024-05-17 16:45:04 发布

郑龙源

最新推荐文章于 2024-05-17 16:45:04 发布

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分类专栏：编程题目

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本文介绍了一个编程问题的解决方法：将一个非负整数的各位数字之和用英文单词形式输出。该解决方案使用了C++语言，并通过字符串处理实现了大整数的计算。

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1005. Spell It Right (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Given a non-negative integer N, your task is to compute the sum of all the digits of N, and output every digit of the sum in English.

Input Specification:

Each input file contains one test case. Each case occupies one line which contains an N (<= 10¹⁰⁰).

Output Specification:

For each test case, output in one line the digits of the sum in English words. There must be one space between two consecutive words, but no extra space at the end of a line.

Sample Input:

Sample Output:

one five

一开始我用的int来接受正整数，记得int的范围在32位机上是21亿，题目的正整数很明显超过这个范围，所以失败了好多次！！

#include<iostream>
#include<cstdio>
#include<string.h>
using namespace std;

void trans(int n)
{
    switch(n)
    {
        case 0: printf("zero");break;
        case 1: printf("one");break;
        case 2: printf("two");break;
        case 3: printf("three");break;
        case 4: printf("four");break;
        case 5: printf("five");break;
        case 6: printf("six");break;
        case 7: printf("seven");break;
        case 8: printf("eight");break;
        case 9: printf("nine");break;
    }
}
int main()
{

    int arr[112];
    char n[112];
    while(scanf("%s",n)!=EOF)
    {
        int i=0,sum=0;
        memset(arr,0,sizeof(arr));
        for(int k=0;k<strlen(n);k++)
        {
                sum+=n[k]-'0';
        }


        if(sum==0)
        {
            printf("zero");
        }
        else
        {
            while(sum)
            {
                arr[i] = sum%10;
                sum = sum/10;
                i++;
            }
            for(int j=i-1;j>=0;j--)
            {

                trans(arr[j]);
                if(j!=0)
                {
                    printf(" ");
                }


            }
        }

        printf("\n");
    }

    return 0;
}