## 1653: [Usaco2006 Feb]Backward Digit Sums

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 318  Solved: 239
[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.

4 16

## Sample Output

3 1 2 4

OUTPUT DETAILS:

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

## Source

Silver

f[1]=3+2;

f[2]=2+1;f[1]=3+2+2+1

f[3]=1+4;f[2]=2+1+1+4;f[1]=3+2+2+1+2+1+1+4;

-----------------------------------------------------------------------------

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int a[11];
int f[11];
int main(){
int n,m;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
a[i]=i;
do{
for(int i=1;i<=n;i++)
f[i]=a[i];
for(int i=1;i<n;i++)
for(int j=i;j>=1;j--)
f[j]+=f[j+1];
if(f[1]==m){
printf("%d",a[1]);
for(int i=2;i<=n;i++)
printf(" %d",a[i]);
printf("\n");
return 0;
}
}while(next_permutation(a+1,a+n+1));
return 0;
}

---------------------------------------------------------------------------------