ACM Backward Digit Sums(挑战程序设计竞赛)

Backward Digit Sums

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 3187
64-bit integer IO format: %lld      Java class name: Main

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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

Source

USACO 2006 February Gold & Silver

简单的枚举

#include <iostream>#include <algorithm>#include <cstdio>using namespace std;#define MAX_N 10int n,finalsum;int mat[MAX_N][MAX_N];int num[MAX_N];int cal(){for(int i=0;i<n;i++)mat[0][i]=num[i];for(int i=0;i<n-1;i++){for(int j=0;j<n-i-1;j++)mat[i+1][j]=mat[i][j]+mat[i][j+1];}return mat[n-1][0];}int main(){scanf("%d%d",&n,&finalsum);for(int i=0;i<n;i++)num[i]=i+1;do{int sum=cal();if(sum==finalsum) break;}while(next_permutation(num,num+n));for(int i=0;i<n-1;i++) printf("%d ",num[i]);printf("%d\n",num[n-1]);return 0;}