(POJ 3187) Backward Digit Sums (递推+暴力)

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 4

解题思路:

全排列1-N,如果有一组数据能够递推最终得到M,那么就输出这组数据

递推要用数组实现,我一开始使用队列递推,结果超时了。。。

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <algorithm>
#include <queue>
#include <string.h>
using namespace std;
int N,M;
int perms[20],resPerms[20];
int main() {
	cin >> N >> M;
	for (int i = 0; i < N; ++i) {
		perms[i] = i+1;
	}
	do 
	{
		for (int i = 0; i < N; ++i) {
			resPerms[i] = perms[i];
		}
		for (int i = N-1; i > 0; --i) {
			for (int j = 0; j <= i-1; ++j) {
				resPerms[j] = resPerms[j] + resPerms[j + 1];
			}
		}
		if (resPerms[0] == M) {
			for (int i = 0; i < N; ++i) {
				cout << perms[i];
				if (i < N - 1) cout << " ";
			}
			break;
		}
	} while (next_permutation(perms,perms+N));
	
	system("PAUSE");
	return 0;
}