UVa1554 - Binary Search

The program fragment below performs binary search of an integer number in an array that is sorted in a nondescending order:

Pascal (file "sproc.pas")C (file "sproc.c")

const  MAXN = 10000;var  A: array[0..MAXN-1] of integer;  N: integer;procedure BinarySearch(x: integer);var  p, q, i, L: integer;begin  p := 0;   { Left border of the search  }  q := N-1; { Right border of the search }  L := 0;   { Comparison counter         }  while p <= q do begin    i := (p + q) div 2;    inc(L);    if A[i] = x then begin      writeln('Found item i = ', i,        ' in L = ', L, ' comparisons');      exit    end;    if x < A[i] then      q := i - 1    else      p := i + 1  endend;

#define MAXN 10000int A[MAXN];int N;void BinarySearch(int x){  int p, q, i, L;  p = 0;   /* Left border of the search  */  q = N-1; /* Right border of the search */  L = 0;   /* Comparison counter         */  while (p <= q) {    i = (p + q) / 2;    ++L;    if (A[i] == x) {      printf("Found item i = %d"        " in L = %d comparisons\n", i, L);      return;    }    if (x < A[i])      q = i - 1;    else      p = i + 1;  }}

Before BinarySearch was called, N was set to some integer number from 1 to 10000 inclusive and array A was filled with a nondescending integer sequence.

It is known that the procedure has terminated with the message "Found item i = XXX in L = XXX comparisons" with some known values of i and L.

Your task is to write a program that finds all possible values of N that could lead to such message. However, the number of possible values of N can be quite big. Thus, you are asked to group all consecutive Ns into intervals and write down only first and last value in each interval.

Input 

The input file consists of several datasets. Each datasets consists of a single line with two integers i and L (0 ≤ i < 10000 and 1 ≤ L ≤ 14), separated by a space.

Output 

On the first line of each dataset write the single integer number K representing the total number of intervals for possible values of N. Then K lines shall follow listing those intervals in an ascending order. Each line shall contain two integers A_i and B_i (A_i ≤ B_i) separated by a space, representing first and last value of the interval.

If there are no possible values of N exist, then the output file shall contain the single 0.

Sample Input 

10 3

Sample Output 

412 1217 1829 3087 94

import java.io.BufferedReader;import java.io.InputStreamReader;import java.io.PrintWriter;import java.io.OutputStreamWriter;import java.io.StreamTokenizer;import java.io.IOException;class Main{public StreamTokenizer tokenizer;public PrintWriter cout;public int n, l;public void init(){BufferedReader cin = new BufferedReader(new InputStreamReader(System.in));tokenizer = new StreamTokenizer(cin);cout = new PrintWriter(new OutputStreamWriter(System.out));}public boolean input() throws IOException{tokenizer.nextToken();if (tokenizer.ttype == StreamTokenizer.TT_EOF) return false;n = (int)tokenizer.nval;tokenizer.nextToken();l = (int)tokenizer.nval;return true;}public boolean check(int x){int p = 0, q = x - 1, m;int cnt = 0;while (p <= q) {m = (p + q) >> 1;cnt++;if (cnt > l) return false;if (m == n) return cnt == l;if (m < n) p = m + 1;else q = m - 1;}return false;}public void solve(){int[] left = new int[10001], right = new int[10001];int cnt = 0;boolean flag = false;for (int i = n + 1; i < 10001; i++) {if (check(i)) {if (!flag) {left[cnt] = i;flag = true;}} else {if (flag) {right[cnt++] = i - 1;flag = false;}}}if (flag) right[cnt++] = 10000;cout.println(cnt);for (int i = 0; i < cnt; i++) {cout.println(left[i] + " " + right[i]);}cout.flush();}public static void main(String[] args) throws IOException{Main solver = new Main();solver.init();while (solver.input()) {solver.solve();}}}