Backward Digit Sums
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 4498 | Accepted: 2586 |
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:
Write a program to help FJ play the game and keep up with the cows.
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.
There are other possible sequences, such as 3 2 1 4, but 3 1 2 4 is the lexicographically smallest.
题目翻译：给出数字n和sum，n代表1-n的n个数字，将1-n按照某种序列排出，然后按照图片的运算方式，就出最后的结果，如果运算结果
等于sum，则输出此时的序列！
解题思路：利用STL里面的next_permutation函数，进行排序，然后一一运算，对比结果！
#include <iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int main()
{
int a[10],b[10],i,j,n,s;
while(scanf("%d%d",&n,&s)==2){
for(i=0;i<n;i++)
a[i]=i+1;
do{
for(i=0;i<n;i++)
b[i]=a[i];
for(i=1;i<n;i++)
for(j=0;j<n-i;j++)
b[j]+=b[j+1];
if(b[0]==s)
break;
}while(next_permutation(a,a+n));
for(i=0;i<n-1;i++)
printf("%d ",a[i]);
printf("%d\n",a[i]);
}
return 0;
}
