zoukankan      html  css  js  c++  java
  • 1745 Divisibility

    Divisibility

    Time Limit: 1000MS Memory Limit: 10000K
    Total Submissions: 14084 Accepted: 4989
    Description

    Consider an arbitrary sequence of integers. One can place + or - operators between integers in the sequence, thus deriving different arithmetical expressions that evaluate to different values. Let us, for example, take the sequence: 17, 5, -21, 15. There are eight possible expressions: 17 + 5 + -21 + 15 = 16
    17 + 5 + -21 - 15 = -14
    17 + 5 - -21 + 15 = 58
    17 + 5 - -21 - 15 = 28
    17 - 5 + -21 + 15 = 6
    17 - 5 + -21 - 15 = -24
    17 - 5 - -21 + 15 = 48
    17 - 5 - -21 - 15 = 18
    We call the sequence of integers divisible by K if + or - operators can be placed between integers in the sequence in such way that resulting value is divisible by K. In the above example, the sequence is divisible by 7 (17+5±21-15=-14) but is not divisible by 5.

    You are to write a program that will determine divisibility of sequence of integers.
    Input

    The first line of the input file contains two integers, N and K (1 <= N <= 10000, 2 <= K <= 100) separated by a space.
    The second line contains a sequence of N integers separated by spaces. Each integer is not greater than 10000 by it’s absolute value.
    Output

    Write to the output file the word “Divisible” if given sequence of integers is divisible by K or “Not divisible” if it’s not.
    Sample Input

    4 7
    17 5 -21 15
    Sample Output

    Divisible
    Source

    Northeastern Europe 1999

    这是简单DP水题,WA了几次,初学的时候这个题还是有必要看看!!

    #include <cstdio> 
    #include <iostream>
    #include <cstring>
    #include <algorithm>
    #include <map>
    #include <cstdlib>
    using namespace std;
    #define mod(a) (a)<0? (-(a))%k:(a)%k  
    #define mem(a,b) memset((a),(b),sizeof(a));
    int dp[10005][105];	
    	int a[10005];
    
    int main(){
    //	freopen("test.txt","r",stdin);
    	mem(dp,0);
    	int n,k;
    	cin>>n>>k;
    	for(int i=1;i<=n;i++)
    		cin>>a[i];
    	dp[1][mod(a[1])]=1;
    	for(int i=2;i<=n;i++)
    		for(int j=0;j<k;j++)
    			if(dp[i-1][j])
    			{
    				dp[i][mod(j+a[i])]=1;
    				dp[i][mod(j-a[i])]=1;
    			}
    	if(dp[n][0]) cout<<"Divisible";
    	else cout<<"Not divisible";
        return 0;}
    
  • 相关阅读:
    cad怎么样创建动态块
    块定义从一个图形传到当前图形
    AutoCAD200X\Support\acad.lsp 启动时自动加载dll
    菜单变灰
    CAD实体双击弹出自定义窗体,可根据扩展数据(通用)
    转载:双击实体弹出对话框(重载AcDbDoubleClickEdit)
    当前已保存的用户坐标系坐标点到世界坐标系的转换
    Windows下编译PHP的C扩展
    flexpaper使用介绍
    IE6 png处理
  • 原文地址:https://www.cnblogs.com/lunatic-talent/p/12798996.html
Copyright © 2011-2022 走看看