henix

#include <assert.h>
#include <stdio.h>
#include <string.h>

typedef int T;

/**
 * [Sorting network](http://en.wikipedia.org/wiki/Sorting_network)
 */
void sort8(T ar[], int n) {

henix

#!/usr/bin/python2
import sys
import re

records = []
for line in sys.stdin:
	records.append(re.match(r'[ ]*(\d)+ ([^ ]+)(.*)', line.rstrip()).groups())

for record in sorted(records, key=lambda d:int(d[0])*(len(d[1])+len(d[2])), reverse=True):
	print "%s %s%s" % record

henix

inline void putint(int n)
{
	static char buf[20];
	register int pos;
	register int x = n;
	if (x == 0) {
		putchar('0');
		return;
	}
	if (x == INT_MIN) { // x = -x do not work for the minimal value of int, so process it first

henix

int getint(char end) {
	int s = 0;
	int ch;
	ch = getchar();
	while (ch != end && ch != EOF) {
		s = s * 10 + ch - '0';
		ch = getchar();
	}
	return s;
}

henix

static inline void putllong(long long n)
{
	static char buf[20];
	register int pos;
	register long long x = n;
	if (x == 0) {
		putchar('0');
		return;
	}
	if (x == LLONG_MIN) { // x = -x do not work for the minimal value of long long, so process it first

henix

typedef int T;

inline bool less(T a, T b) {
	return a < b;
}

inline bool lessEq(T a, T b) {
	return a <= b;
}

henix

int getint2(int *out) {
	register int s = 0;
	register int ch;
	ch = getchar();
	bool negative = false;
	while (ch < '0' || ch > '9') {
		if (ch == '-') {
			negative = true;
			ch = getchar();
			break;

henix

/*************
* colors.js *
 *************
 *
 * You're almost at the exit. You just need to get past this
 * color lock.
 *
 * Changing your environment is no longer enough. You must
 * learn to change yourself. I've sent you a little something
 * that should help with that.

henix

rel() {
	[ ! -d 'target/universal/stage' ] && echo stage not exist! && return 1
	name=$(pwd | tr '/' '\n' | tail -n1)
	timestamp=$(date +%Y%m%d_%H%M%S)
	cp -r "target/universal/stage" "/apps/$name/stage-$timestamp"
	ln -sTv "stage-$timestamp" "/apps/$name/stage"
}

henix

有一存有整数的数组 A[n] (n <= 2^31 - 1, 但需要的话可以再扩大), 其中 0 <= i, A[i] <= n-1 且 i!=j 时 A[i]!=A[j] (即一个排列). 用你认为*最高效*的方法求 B=A^(-1) (即对范围内满足 A[B[i]]=i 的 B)