Problem description

Despite the glorious fall colors in the midwest, there is a great deal of time to spend while on a train from St. Louis to Chicago. On a recent trip, we passed some time with the following game.

We start with a positive integer S. So long as it has more than one digit, we compute the product of its digits and repeat. For example, if starting with 95, we compute 9 × 5 = 45 . Since 45 has more than one digit, we compute 4 × 5 = 20 . Continuing with 20, we compute 2 × 0 = 0 . Having reached 0, which is a single-digit number, the game is over.

As a second example, if we begin with 396, we get the following computations:

3 × 9 × 6 = 162

1 × 6 × 2 = 12

1 × 2 = 2

and we stop the game having reached 2.

Input
   Each line contains a single integer 1 ≤ S ≤ 100000, designating the starting value. The value S will not have any leading zeros. A value of 0 designates the end of the input.
Output
  For each nonzero input value, a single line of output should express the ordered sequence of values that are considered during the game, starting with the original value.
Sample Input
95
396
28
4
40
0
Sample Output

95 45 20 0396 162 12 228 16 6440 0

题意:给出一个数字,将每一位相乘得到下一个数字,知道数字位数为1则停止,输出所有情况

水题,不解释

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std; int main()
{
int n,t,r,s;
while(~scanf("%d",&n),n)
{
int cnt = 0;
printf("%d",n);
if(n>=10)
{
while(n)
{
t = n;
s = 1;
while(t)
{
r = t%10;
s*=r;
t/=10;
}
n = s;
if(n/10==0)
{
printf(" %d",n);
break;
}
printf(" %d",s);
}
}
printf("\n");
} return 0;
}

HUNNU11352:Digit Solitaire的更多相关文章

  1. 四校训练 warm up 14

    A:Pythagoras's Revenge 代码: #include<cstdio> #define ll long long using namespace std; int main ...

  2. [LeetCode] Nth Digit 第N位

    Find the nth digit of the infinite integer sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... Note: n i ...

  3. [LeetCode] Number of Digit One 数字1的个数

    Given an integer n, count the total number of digit 1 appearing in all non-negative integers less th ...

  4. [Leetcode] Number of Digit Ones

    Given an integer n, count the total number of digit 1 appearing in all non-negative integers less th ...

  5. 【Codeforces715C&716E】Digit Tree 数学 + 点分治

    C. Digit Tree time limit per test:3 seconds memory limit per test:256 megabytes input:standard input ...

  6. kaggle实战记录 =>Digit Recognizer

    date:2016-09-13 今天开始注册了kaggle,从digit recognizer开始学习, 由于是第一个案例对于整个流程目前我还不够了解,首先了解大神是怎么运行怎么构思,然后模仿.这样的 ...

  7. [UCSD白板题] The Last Digit of a Large Fibonacci Number

    Problem Introduction The Fibonacci numbers are defined as follows: \(F_0=0\), \(F_1=1\),and \(F_i=F_ ...

  8. Last non-zero Digit in N!(阶乘最后非0位)

    Last non-zero Digit in N! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Jav ...

  9. POJ3187Backward Digit Sums[杨辉三角]

    Backward Digit Sums Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6350   Accepted: 36 ...

随机推荐

  1. 移动端网页JS框架-手机触摸事件框架,日历框架带滑动效果

    swiper.js,hammer.js,mobiscroll http://www.mobiscroll.com/       日历

  2. sql语句查询添加自增列

    SELECT Row_Number() over ( order by getdate() ) as '序号', * FROM  T_Cod_XQ

  3. 百度地图 javascript相关Bug搜集

    一 在手机里用百度地图js版做webapp   bug集合 1 之前用2.0版本的时候发现只要地图添加了覆盖物,无论数量多少,当地图放大到很小的范围时候,会卡死 1.1 当时处理办法:将版本降低至1. ...

  4. c 有意思的数组初始化

    c 有意思的数组初始化 #include <stdio.h> int main() { int i = 0; char a[1024]; char a0[10] = {}; char a1 ...

  5. hadoop format过程

    private static boolean format(Configuration conf, boolean isConfirmationNeeded ) throws IOException ...

  6. VC 中与字符串相关的宏 _T、TEXT,_TEXT、L 的作用(简单明了)

    一. 在字符串前加一个L作用:    如  L"我的字符串"    表示将ANSI字符串转换成unicode的字符串,就是每个字符占用两个字节.   strlen("as ...

  7. Java Web Services (0) - Overview

    前言第1章 Web服务快速入门 1.1 Web服务杂项 1.2 Web服务有什么好处 1.3 Web服务和面向服务的架构 1.4 Web服务简史 1.4.1 从DCE/RPC到XML-RPC 1.4. ...

  8. Lambda高手之路第二部分

    转http://www.cnblogs.com/lazycoding/archive/2013/01/06/2847579.html 闭包的影响 为了展示闭包的影响,我们看下面这个例子. var bu ...

  9. 数字雨Shopex 4.8.5 SQL Injection Exp

    # -*- coding:utf-8 -* #Author:MXi4oyu #Email:798033502@qq.com #Shopex 4.8.5 SQL Injection Exp #转载请说明 ...

  10. Android ARM汇编语言

    简介 ARM是Advanced RISC Machine的首字母缩写,它可以称之为一家嵌入式处理器的提供商,也可以理解为一种处理器的架构,还可以将它作为一套完整的处理器指令集. 原生程序与ARM汇编语 ...