zoukankan      html  css  js  c++  java
  • poj 3187 Backward Digit Sums

    Backward Digit Sums
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions:   Accepted:

    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.


    AC代码例如以下:



    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    
    int cont=0;
    int n,i,a[15],vis[15],sum,bj=0;
    void yz()
    {
        int i,j,b[15];
        for(i=0;i<n;i++)
            b[i]=a[i];
        for(i=n-2;i>=0;i--)
            for(j=0;j<=i;j++)
        {
            b[j]+=b[j+1];
        }
        if(b[0]==sum)
            bj=1;
    }
    
    
    int main()
    {
        int i;
        scanf("%d%d",&n,&sum);
        memset(vis,0,sizeof(vis));
        for(i=0;i<n;i++)
            a[i]=i+1;
        do
        {
            yz();
            if(bj==1)
            {
                for(i=0;i<n-1;i++)
                    printf("%d ",a[i]);
                printf("%d
    ",a[i]);
                break;
            }
        }
        while(next_permutation(a,a+n));
    
    
        return 0;
    
    }
    


  • 相关阅读:
    PAT Advanced 1008 Elevator (20) [数学问题-简单数学]
    PAT Advanced 1051 Pop Sequence (25) [栈模拟]
    PAT Basic 完美数列(25) [two pointers]
    PAT Basic 插⼊与归并(25) [two pointers]
    PAT Advanced 1089 Insert or Merge (25) [two pointers]
    第五章 数据的共享和保护
    第四章 面向对象程序设计的基本特点 课堂笔记
    第三章 函数 课堂笔记
    扫描线-Meteor UVALive
    贪心-Stall Reservations POJ
  • 原文地址:https://www.cnblogs.com/wzjhoutai/p/6727869.html
Copyright © 2011-2022 走看看