poj 3178 Backward Digit Sums (全排列)

最新推荐文章于 2020-01-25 19:50:32 发布

阿.荣.

最新推荐文章于 2020-01-25 19:50:32 发布

阅读量524

点赞数

CC 4.0 BY-SA版权

分类专栏： STL 其它oj

本文链接：https://blog.youkuaiyun.com/bmicnj/article/details/51150866

其它oj 同时被 2 个专栏收录

64 篇文章

订阅专栏

STL

30 篇文章

订阅专栏

本文介绍了一款逆向思维的数字游戏，玩家需根据最终总和及数字范围，找出初始数字序列。通过全排列算法及next_permutation()函数的应用，实现对所有可能序列的遍历，最终找到符合条件的最小字典序解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

<div class="ptt" lang="en-US" style="text-align: center; font-size: 18pt; font-weight: bold; color: blue;">Backward Digit Sums</div><div class="plm" style="text-align: center;font-size:14px;"><table align="center"><tbody><tr><td><strong>Time Limit:</strong> 1000MS</td><td width="10px"> </td><td><strong>Memory Limit:</strong> 65536K</td></tr><tr><td><strong>Total Submissions:</strong> 5858</td><td width="10px"> </td><td><strong>Accepted:</strong> 3398</td></tr></tbody></table></div><p class="pst" style="font-size: 18pt; font-weight: bold; color: blue;">Description</p><div class="ptx" lang="en-US" style="font-family: 'Times New Roman', Times, serif;font-size:14px;">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: 

<pre>    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

再次用到了全排列，next_permutation()的用法！STL！

代码：

<pre name="code" class="html">#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

int main()
{
    int n,s;
    int a[20],b[20];
    while(~scanf("%d%d",&n,&s))
    {
        for(int i=0;i<n;i++)
            a[i]=b[i]=i+1;
        do
        {
            for(int i=0;i<n;i++)
                a[i]=b[i];
            for(int j=n;j>0;j--)
            {
                for(int k=0;k<j-1;k++)
                {
                    a[k]=a[k]+a[k+1];
                }
            }
            if(a[0]==s)
            {
                for(int i=0;i<n-1;i++)
                    printf("%d ",b[i]);
                printf("%d\n",b[n-1]);
                break;
            }
        }while(next_permutation(b,b+n));
    }
    return 0;
}