Nyoj 1057 Find maximum number (iii) (knowledge of the greedy algorithm)

Before doing greedy topics are the first step to the data from the large to the small sort, most of the greedy problem of the purpose of the treatment method is also the case. But Nyoj's 1057 questions, in solving the problem

do not need to sort, not realize at a time is greedy problem. After reading the discussion area, I realized that we should use greedy thoughts to solve the problem. It is realized that the greedy algorithm is not the mechanical operation after sorting, but the idea of seeking the optimal solution from the local.

The original question is as follows:

Find the maximum number (c) time limit: +Ms | Memory Limit:65535KB Difficulty:2

Describe

Given an integer n, you can move a number on 2 adjacent digits at a time, moving up to K times, and getting a new integer.

What is the maximum value for this new integer.

Input

multiple sets of test data.
One row for each set of test data, with two numbers n and K per line (1?≤? n≤?10^18; 0?≤? K?≤?100).

Output

The

output of each set of test data takes one row, and the output moves to get the maximum value of the new integer.

Sample input

1990 1100 09090000078001234 6

Sample output

91901009907000008001234

The code is as follows:

#include <stdio.h> #include <stdlib.h> #include <string.h>int main () {int N,t,i,k,len;char max,str[ 100];while (scanf ("%s%d", str,&k)!=eof) {Len=strlen (str); t=0;//t mark the first position of each exchange start while (k>0) {if (T==len)  break;n=t;//the first character for the interchange portion with n marks at the beginning of each interchange for (i=t;i<=t+k;i++) {if (I>=len)  break;if (Str[i]>str[n])  n=i ;//When you encounter a character with a large number of characters, mark it with N. }if (n!=t) {max=str[n];for (i=n;i>t;i--)  str[i]=str[i-1];str[t]=max;k-=n-t;//equals k=k-(N-T);} t++;} printf ("%s\n", str);} return 0;}

Nyoj 1057 Find maximum number (iii) (knowledge of the greedy algorithm)