POJ - 3187 Backward Digit Sums

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. 

题意：就是给你一个n和sum,wenni 1~n的这些数中，怎么排列，能通过如上图的方式得到的和与sum相同。

思路：当是A 就得A   

若是 A B         若是   A      B       C                                 A            B              C              D

        A+B                    A+B   B+C                                        A+B        B+C       C+D  

                                    A+B+B+C                                         A+B+B+C        B+C+C+D

                                                                                                 A+B+B+B+C+C+C+D

依次类推，得到几个数就分别乘以第几层的杨辉三角

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1

#include
#include
#include
#include
using namespace std;
int main(){
    int a[11],b[11];
    int y[11][11];
    int n,sum,i,j;
    memset(y,0,sizeof(y));
    y[1][1]=1;
    for(i=2;i<=10;i++){
        for(j=1;j<=i;j++){
          y[i][j]=y[i-1][j]+y[i-1][j-1];
        }
    }
  /*
    for(i=1;i<=10;i++){
        for(j=1;j<=i;j++){
          printf("%d ",y[i][j]);
        }
        printf("\n");
    }
*/
    scanf("%d%d",&n,&sum);
    for(i=1;i<=n;i++)
        a[i]=i;

    do{
        int sum1=0;
       for(i=1;i<=n;i++){
        sum1+=a[i]*y[n][i];
       }
       if(sum1==sum){
        break;
       }

    }while(next_permutation(a+1,a+n+1));//排列组合
    for(i=1;i<=n;i++){
        if(i!=n)
            printf("%d ",a[i]);
        else
            printf("%d\n",a[i]);
    }
}