POJ1350模拟

Cabric Number Problem

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11008 Accepted: 3379

Description

If we input a number formed by 4 digits and these digits are not all of one same value, then it obeys the following law. Let us operate the number in the following way: 
(1) Arrange the digits in the way from bigger to smaller, such that it forms the biggest number that could be made from these 4 digits; 
(2) Arrange the digits in the way from smaller to bigger, such that it forms the smallest number that could be made from these 4 digits (If there is 0 among these 4 digits, the number obtained may be less than four digits); 
(3) Find the difference of these two numbers that is a new four digital number. 
Repeat the above process, we can finally always get the result 6174 or 0. 
Please write the program to realize the above algorithm. 

Input

Each case is a line of an integer.-1 denotes the end of input.

Output

If the integer is formed exactly by 4 digits and these digits are not all of one same value, then output from the program should show the procedure for finding this number and the number of repetition times. Otherwise output "No!!".

Sample Input

536422214444-1

Sample Output

N=5364:6543-3456=30878730-378=83528532-2358=6174Ok!! 3 timesN=2221:2221-1222=999999-999=0Ok!! 2 timesN=4444:No!!

Source

Xi'an 2002

    题目滴意思看输入输出样例便可知道，即一个四位数，如果是四位全相同，直接输出No!!，不用进行后续运算。如果四位不全相同，用它的各位组成的最大数减去最小数，这个差不为0或者6174时，打印出这个算式，继续用这个差来进行前面的操作，最后输出进行运算的次数。没什么别的方法，开循环运行一次做一次判断，直到满足条件跳出。主要是位数的问题。还有一点值得注意，这个题目输入的数据如果不是四位数，要输出四位相等时的输出，本人在这里WA两次。下面是我的代码

#include <stdio.h>#include <math.h>#include <stdlib.h>#include <iostream>using namespace std;int cmp(const void*a,const void*b){    return *(int *)a-*(int *)b;}int sum1(int a[],int n){    if(n==4) return 1000*a[0]+100*a[1]+10*a[2]+a[3];    else if(n==3) return 100*a[0]+10*a[1]+a[2];    else if(n==2) return 10*a[0]+a[1];    else if(n==1) return a[0];}int sum2(int a[],int n){    if(n==4) return 1000*a[3]+100*a[2]+10*a[1]+a[0];    else if(n==3) return 100*a[2]+10*a[1]+a[0];    else if(n==2) return 10*a[1]+a[0];    else if(n==1) return a[0];}int main(){    int number;    //freopen("in.txt","r",stdin);    while(~scanf("%d",&number)&&number!=-1)    {        int a[4],digit1,digit2,s,ss,numbers,times=0,t=4;        numbers=number;        cout<<"N="<<number<<":"<<endl;        if(number<=999||number>=10000)  /*不满足四位数跳出*/        {            printf("No!!\n");            continue;        }        for(int i=0; i<4; i++)        {            a[i]=number%10;            number/=10;        }        if(a[0]==a[1]&&a[1]==a[2]&&a[2]==a[3])  /*全相等的时候直接输出N哦！！跳出*/        {            printf("No!!\n");            continue;        }        do        {            int b[4];            for(int i=0; i<t; i++)            {                b[i]=numbers%10;                numbers/=10;            }            qsort(b,t,sizeof(int),cmp);     /*排序*/            digit1=sum2(b,t);            digit2=sum1(b,t);            s=digit1-digit2;            printf("%d-%d=%d\n",digit1,digit2,s);            ss=numbers=s;            times++;            for(t=0; ss!=0; t++) ss/=10;     /*t用来存位数*/        }        while(s!=0&&s!=6174);        printf("Ok!! %d times\n",times);    }    return 0;}