POJ 题目Backward Digit Sums(next_permutation)

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本文详细介绍了BackwardDigitSums问题的解决策略，包括算法设计、实现细节及优化技巧。通过实例演示了如何从最终总和逆向推算原始数列，旨在帮助读者理解并掌握该算法的核心思想。

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Backward Digit Sums

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4807		Accepted: 2772

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 4

Hint

Explanation of the sample:

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

Source

USACO 2006 February Gold & Silver

ac代码

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
int a[15],b[15];
int main()
{
	int n,m;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		int i,j;
		for(i=1;i<=n;i++)
		{
			a[i]=i;
		}
		do{
			for(i=1;i<=n;i++)
				b[i]=a[i];
			for(i=1;i<=n;i++)
			{
				for(j=1;j<=n-i;j++)
					b[j]+=b[j+1];
			}
			if(b[1]==m)
				break;
			next_permutation(a+1,a+n+1);
		}while(1);
		for(i=1;i<=n;i++)
		{
			printf("%d%c",a[i],i==n?'\n':' ');
		}
	}
}