pat basic level 1019

题目地址：http://pat.zju.edu.cn/contests/pat-b-practise/1019

Github:https://github.com/kunth/evernote/blob/master/pat_basic_level_1019

//Date : 2013-11-05
//Author : fibonacci
//Note: http://pat.zju.edu.cn/contests/pat-b-practise/1019 //Acceptd
//此题要注意输出格式

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;

int main()
{
int N, a, b, c, d, num, revnum;
vector<int> vec;
//freopen("in.txt", "r", stdin);
scanf("%d", &N);
a = N / 1000;
b = (N % 1000)/100;
c = (N % 100)/10;
d = N % 10;
if(a==b && b==c && c==d)
{
printf("%04d - %04d = 0000\n", N, N);
return 0;
}
vec.push_back(a);
vec.push_back(b);
vec.push_back(c);
vec.push_back(d);
sort(vec.begin(), vec.end());
num = vec.at(3) * 1000 + vec.at(2)*100 + vec.at(1)*10 +vec.at(0);
revnum = vec.at(0) * 1000 + vec.at(1)*100 + vec.at(2)*10 +vec.at(3);
printf("%04d - %04d = %04d\n", num, revnum, num-revnum);
N = num-revnum;
while(N!=6174)
{
vec.clear();
a = N / 1000;
b = (N % 1000)/100;
c = (N % 100)/10;
d = N % 10;
vec.push_back(a);
vec.push_back(b);
vec.push_back(c);
vec.push_back(d);
sort(vec.begin(), vec.end());
num = vec.at(3) * 1000 + vec.at(2)*100 + vec.at(1)*10 +vec.at(0);
revnum = vec.at(0) * 1000 + vec.at(1)*100 + vec.at(2)*10 +vec.at(3);
printf("%04d - %04d = %04d\n", num, revnum, num-revnum);
N = num - revnum;
}

return 0;