Pasha Maximizes

题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=82662#problem/E

标准的模拟题：总共可以交换k次，那就在k次以内找能进行交换的最小的数字，然后，用k减去消耗的次数

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
#define ll __int64

int main()
{
    char a[2000];
    ll k;
    while(cin >> a >> k)
    {
        int flag = -1;
        while(k)
        {
            flag++;
            char Max = a[flag];
            int temp = -1;
            for(int i = 1; i <= k; i++)
            {
                if(a[i+flag]>Max)
                {
                    Max = a[i+flag];
                    temp = i;
                }
            }
            //cout << flag << endl;
            //cout << Max << " " << temp << endl;
            for(int m = temp+flag; m > flag; m--)
            {
                swap(a[m],a[m-1]);
            }
           if(temp!=-1)
                k -= temp;
            //cout << k << endl;
        }
        puts(a);
    }
    return 0;
}