Num 27:nyoj:0448 looking for maximum number [greedy]

A classic greedy question: title:

Find the maximum number of time limits: +Ms | Memory Limit:65535KB Difficulty:2

Describe

Delete the M numbers in the integer n so that the remaining digits are the largest of the new numbers in the original order,

For example, when n=92081346718538,m=10, the new maximum number is 9888.

Input

The first line enters a positive integer t, which indicates that there is a T group of test data
One row for each set of test data, with two numbers per row n,m (n may be a large integer, but the number of bits does not exceed 100 bits, and the number of digits of the first non-0,m less than the integer n is guaranteed)

Output

The output of each set of test data is one row, the maximum new number of the output remaining number in the original order

Sample input

292081346718538 101008908 5

Sample output

988898

Topic Analysis: Delete a certain number of numbers from a series of values, make the remaining number the largest;

Problem Solving Ideas:

first find the maximum value from s[0]-s[m] and record the next ID.
The second time, from the position of id+1, find the maximum value between s[id+1]-s[m+1] and record the position.
By this method, the end of the string is always found, that is, m==lens-1.

Code implementation:

1. Integers of more than 100 digits, implying that we want to use a character array to store;

2. The first number must be the largest number in front of the data string (in fact, each number from the back is the largest of the small range that starts back from this number);

AC Code:

#include <stdio.h> #include <string.h>char a[110];int main () {int t;scanf ("%d", &t), while (t--) {int Len, I,m,p;char ch;scanf ("%s%d", &a,&m); Len=strlen (a);p =0;while (M<len) {ch=a[p];for (i=p;i<=m;i++) if (A[i] >a[p]) {ch=a[i];p =i;} printf ("%c", ch); m++;p + +;} printf ("\ n");} return 0;}

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Num 27:nyoj:0448 looking for maximum number [greedy]