BZOJ 1653: [Usaco2006 Feb]Backward Digit Sums(暴力)

1653: [Usaco2006 Feb]Backward Digit Sums

Time Limit: 5 Sec  Memory Limit: 64 MB Submit: 349  Solved: 258 [Submit][Status][Discuss]

Description

FJ and his cows enjoy playing a mental game. They write down the numbers from 1 to N (1 <= N <= 10) in a certain order and then sum adjacent numbers to produce a new list with one fewer number. They repeat this until only a single number is left. For example, one instance of the game (when N=4) might go like this: 3 1 2 4 4 3 6 7 9 16 Behind FJ's back, the cows have started playing a more difficult game, in which they try to determine the starting sequence from only the final total and the number N. Unfortunately, the game is a bit above FJ's mental arithmetic capabilities. Write a program to help FJ play the game and keep up with the cows.

Input

• Line 1: Two space-separated integers: N and the final sum.

Output

• Line 1: An ordering of the integers 1..N that leads to the given sum. If there are multiple solutions, choose the one that is lexicographically least, i.e., that puts smaller numbers first.

Sample Input

4 16

Sample Output

3 1 2 4OUTPUT DETAILS:

There are other possible sequences, such as 3 2 1 4, but 3 1 2 4 is the lexicographically smallest.

思路：n很小。。一开始做得时候把公式推错了。。3+3算成了4.我的内心是崩溃的。。

其实直接暴力就好。可以用next_permutation来生成全排列，然后判断是否合法。

/* ***********************************************
Author :111qqz
Created Time :2016年04月12日 星期二 10时54分23秒
File Name :code/bzoj/1653.cpp
************************************************ */

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair

using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
int n;
int sum;
int a[15];
int tmp[15];
int main()
{
	#ifndef  ONLINE_JUDGE 
	freopen("code/in.txt","r",stdin);
  #endif

	scanf("%d %d",&n,&sum);
	for ( int i = 1 ; i <= n ; i++) a[i] = i;
	
	do
	{
	    for ( int i = 1 ; i <= n ; i++) tmp[i] = a[i];
	    for ( int i = 1 ; i < n ; i++)
		for ( int j = 1 ; j <= n - i ; j ++)
		    tmp[j]+=tmp[j+1];
	    if (tmp[1]==sum)
	    {
		for ( int i = 1 ; i< n ; i++)
		    printf("%d ",a[i]);
		printf("%d\n",a[n]);
		break;
	    }
	}while (next_permutation(a+1,a+n+1));


  #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}