One measure of ``unsortedness‘‘ in a sequence is the number of pairs of entries that are out of order with respect to each other. For instance, in the letter sequence ``DAABEC‘‘, this measure is 5, since D is greater than four letters to its right and E is greater than one letter to its right. This measure is called the number of inversions in the sequence. The sequence ``AACEDGG‘‘ has only one inversion (E and D)---it is nearly sorted---while the sequence ``ZWQM‘‘ has 6 inversions (it is as unsorted as can be---exactly the reverse of sorted). 

You are responsible for cataloguing a sequence of DNA strings (sequences containing only the four letters A, C, G, and T). However, you want to catalog them, not in alphabetical order, but rather in order of ``sortedness‘‘, from ``most sorted‘‘ to ``least sorted‘‘. All the strings are of the same length. 

The first line contains two integers: a positive integer n (0 < n <= 50) giving the length of the strings; and a positive integer m (0 < m <= 100) giving the number of strings. These are followed by m lines, each containing a string of length n.

10 6AACATGAAGGTTTTGGCCAATTTGGCCAAAGATCAGATTTCCCGGGGGGAATCGATGCAT

CCCGGGGGGAAACATGAAGGGATCAGATTTATCGATGCATTTTTGGCCAATTTGGCCAAA

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <stack>#include <vector>#include <queue>#include <cstring>#include <string>#include <map>#include <set>using namespace std;const int BufferSize = 1 << 16;char buffer[BufferSize], *Head, *Tail;inline char Getchar() {    if(Head == Tail) {        int l = fread(buffer, 1, BufferSize, stdin);        Tail = (Head = buffer) + l;    }    return *Head++;}int read() {    int x = 0, f = 1; char c = getchar();    while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); }    while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); }    return x * f;}#define maxn 110#define maxl 60int n, l;char S[maxn][maxl];vector <int> id[maxn*maxn];void process(int x) {	int A = 0, C = 0, G = 0, sum = 0;	for(int i = l; i >= 1; i--) {		if(S[x][i] == ‘A‘) A++;		if(S[x][i] == ‘C‘) sum += A, C++;		if(S[x][i] == ‘G‘) sum += A + C, G++;		if(S[x][i] == ‘T‘) sum += A + C + G;	}	id[sum].push_back(x);	return ;}int main() {	l = read(); n = read();	for(int i = 1; i <= n; i++) scanf("%s", S[i] + 1), process(i);		for(int i = 0; i <= l * l; i++)		for(int j = 0; j < id[i].size(); j++) printf("%s\n", S[id[i][j]] + 1);		return 0;}