atcoder ---ID

Problem Statement

In Republic of Atcoder, there are N prefectures, and a total of M cities that belong to those prefectures.

City i is established in year Yi and belongs to Prefecture Pi.

You can assume that there are no multiple cities that are established in the same year.

It is decided to allocate a 12-digit ID number to each city.

If City i is the x-th established city among the cities that belong to Prefecture i, the first six digits of the ID number of City i is Pi, and the last six digits of the ID number is x.

Here, if Pi or x (or both) has less than six digits, zeros are added to the left until it has six digits.

Find the ID numbers for all the cities.

Note that there can be a prefecture with no cities.

Constraints

• 1≤N≤105
• 1≤M≤105
• 1≤Pi≤N
• 1≤Yi≤109
• Yi are all different.
• All values in input are integers.

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Input

Input is given from Standard Input in the following format:

N M
P1 Y1
:
PM YM

Output

Print the ID numbers for all the cities, in ascending order of indices (City 1, City 2, …).

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Sample Input 1

Copy

2 3
1 32
2 63
1 12

Sample Output 1

Copy

000001000002
000002000001
000001000001

• As City 1 is the second established city among the cities that belong to Prefecture 1, its ID number is 000001000002.
• As City 2 is the first established city among the cities that belong to Prefecture 2, its ID number is 000002000001.
• As City 3 is the first established city among the cities that belong to Prefecture 1, its ID number is 000001000001.

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Sample Input 2

Copy

2 3
2 55
2 77
2 99

Sample Output 2

Copy

000002000001
000002000002
000002000003

先整个进行时间的排序，再遍历一遍

#include<cstdio>
#include<algorithm>
using namespace std;
struct ss{
	int rank;
    int p;
    int y;
}a[100005];
int cmp(ss x,ss y){
    if(x.p!=y.p) return x.p<y.p;
    else return x.y<y.y;
}
int main()
{
    int n ,m;
    scanf("%d%d",&n,&m);
    int b[100005];
    for(int i=0;i<m;i++){
    	a[i].rank=i;
        scanf("%d%d",&a[i].p,&a[i].y); 	
        b[i]=a[i].p;
	}
    sort(a,a+m,cmp);
    int f=a[0].p,flag=1;
    int x[100005];
    
    for(int i=0;i<m;i++){
    	if(a[i].p!=f) {flag=1;f=a[i].p;}
    	x[a[i].rank]=flag++;
	}
    // for(int i=0;i<m;i++)
    //    cout<< a[i].rank<<" "<<a[i].p<<" " <<a[i].y<<endl;
    for(int i=0;i<m;i++)
        printf("%06d%06d\n",b[i],x[i]);
    return 0;
}