POJ3187 Backward Digit Sums 全排列经典题目

Backward Digit Sums

Time Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

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

题意：根据题意，给出一共n行，还有最后一行的数是m。依题意得，第一行一共n个数，而且每个数还不能相等。

输出满足条件的最小的排列，考察next_permutation的用法

#include <iostream>#include <algorithm>#include <stdio.h>#include <string.h>using namespace std;int a[111][11];int s[111];int main(){    int n,m;    while(~scanf("%d%d",&n,&m))    {        for(int i = 1;i <= n;i ++)            s[i] = i;        do        {            memset(a,0,sizeof(a));            for(int i = 1;i <= n;i ++)                a[n][i] = s[i];            int t = n;            for(int i = t - 1;i >= 1;i --)            {                for(int j = 1;j <= i;j ++)                    a[i][j] = a[i+1][j] + a[i+1][j+1];            }            if(a[1][1] == m)            {                for(int i = 1;i < n;i ++)                    printf("%d ",s[i]);                printf("%d\n",s[n]);                break;            }            ///注意s[i]是从第一个位置开始输出的，所以要对应到next_permutation(s+1,s + n+1)        }while(next_permutation(s+1,s + n+1));    }    return 0;}