Ural 1100. Final Standings 桶排序

1100. Final Standings

Time limit: 1.0 second
Memory limit: 16 MB

Old contest software uses bubble sort for generating final standings. But now, there are too many teams and that software works too slow. You are asked to write a program, which generates exactly the same final standings as old software, but fast.

Input

The first line of input contains only integer 1 < N ≤ 150000 — number of teams. Each of the next Nlines contains two integers 1 ≤ ID ≤ 107 and 0 ≤ M ≤ 100. ID — unique number of team, M — number of solved problems.

Output

Output should contain N lines with two integers ID and M on each. Lines should be sorted by M in descending order using bubble sort (or analog).

Sample

inputoutput

81 216 311 220 33 526 47 122 4

3 526 422 416 320 31 211 27 1

Hint

Bubble sort works following way: 
while (exists A[i] and A[i+1] such as A[i] < A[i+1]) do
   Swap(A[i], A[i+1]);

Problem Author: Pavel Atnashev
Problem Source: Tetrahedron Team Contest May 2001 

/** * * 题目要求要改进冒泡，这就要求排序稳定 * m是100以内的整数，可以用桶排序 * */#include <iostream>#include <cstdio>using namespace std;int id[150005],m[150005];int main(){    int i,j,n;    scanf("%d",&n);    for(i=0; i<n; i++)        scanf("%d%d",&id[i],&m[i]);    for(i=100; i>=0; i--)        for(j=0; j<n; j++)            if(m[j]==i)                printf("%d %d\n",id[j],m[j]);    return 0;}