• <ins id="pjuwb"></ins>
    <blockquote id="pjuwb"><pre id="pjuwb"></pre></blockquote>
    <noscript id="pjuwb"></noscript>
          <sup id="pjuwb"><pre id="pjuwb"></pre></sup>
            <dd id="pjuwb"></dd>
            <abbr id="pjuwb"></abbr>
            posts - 195,  comments - 30,  trackbacks - 0

            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

            There are multiple test cases, the first line is the number of test cases.
            The first line of each test case 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

            2
            4 7
            17 5 -21 15
            4 5
            17 5 -21 15
            

            Sample Output

            Divisible
            Not divisible
            
            啟發:1,涉及整除就要聯想取模!!!!!!!!
                        2,這是有層次性,不要搞混,以下是錯誤代碼 
            #include<iostream>
            #include
            <cstdlib>
            using namespace std;
              
              
            int main()
              
            {
              freopen(
            "s.txt","r",stdin);
              freopen(
            "key.txt","w",stdout);
              
            int testnum;
              
            int num,pos;
              cin
            >>testnum;
              
            int a[101],temp,i,j;
              
            while(testnum--)
              
            {
                    memset(a,
            0,sizeof(a));
                    cin
            >>num>>pos;
                    cin
            >>temp;
                    temp
            %=pos;
                    
            if(temp<0)temp+=pos;
                    a[temp]
            =1;    
                    
            for(i=1;i<num;i++)
                    
            {
                        cin
            >>temp;
                        temp
            %=pos;
                        
            if(temp<0)temp+=pos;//temp%pos¿ÉÄÜÊǸºÊý 
                        for(j=0;j<pos;j++)
                        
            {
                            
            if(a[j]>0)
                              
            {
                                    a[(j
            +temp)%pos]++;
                                    
            if((j-temp)<0)
                                    
            {
                                        a[(j
            -temp)+pos]++;
                                    }

                                    
            else
                                    
            {
                                    a[j
            -temp]++;
                                    }

                                    
            if(temp!=0)
                                    a[j]
            =0;
                              }

                        }

                    }

                    
            if(a[0]>0)cout<<"Divisible"<<endl;
                    
            else cout<<"Not divisible"<<endl;
                }


              
            //system("PAUSE");
              return   0;
              }

            這是有層次性,再添加一個元素時,只能改變前一組a[j]的值

             if(a[j]>0)//
                  {
                  a[(j+temp)%pos]++;//這里的修改應該不應添加到前面的a[j]中去。
                        if((j-temp)<0)

            所以應當用兩個數組。

            #include<iostream>
            #include
            <cstdlib>
            using namespace std;
              
              
            int main()
              {
              freopen(
            "s.txt","r",stdin);
              freopen(
            "key.txt","w",stdout);
              
            int testnum;
              
            int num,pos;
              cin
            >>testnum;
              
            int a[101],temp,i,j;
              
            int b[101];
              
            while(testnum--)
              {
                    memset(a,
            0,sizeof(a));
                    memset(a,
            0,sizeof(b));
                    cin
            >>num>>pos;
                    cin
            >>temp;
                    temp
            %=pos;
                    
            if(temp<0)temp+=pos;
                    a[temp]
            =1;    
                    
            for(i=1;i<num;i++)
                    {
                        cin
            >>temp;
                        temp
            %=pos;
                        
            if(temp<0)temp+=pos;//temp%pos¿ÉÄÜÊǸºÊý 
                        for(j=0;j<pos;j++)
                        {
                            
            if(a[j]>0)
                              {
                                    b[(j
            +temp)%pos]++;
                                    
            if((j-temp)<0)
                                    {
                                        b[(j
            -temp)+pos]++;
                                    }
                                    
            else
                                    {
                                    b[j
            -temp]++;
                                    }
                              }
                        }      
                              memset(a,
            0,sizeof(a));
                              
            for(j=0;j<pos;j++)
                              {
                                    
            if(b[j]>0)
                                    a[j]
            =1;
                              }
                              memset(b,
            0,sizeof(b));
                    }
                    
            if(a[0]>0)cout<<"Divisible"<<endl;
                    
            else cout<<"Not divisible"<<endl;
                }

              
            //system("PAUSE");
              return   0;
              }
            posted on 2009-06-30 22:20 luis 閱讀(261) 評論(0)  編輯 收藏 引用
            <2012年4月>
            25262728293031
            1234567
            891011121314
            15161718192021
            22232425262728
            293012345

            常用鏈接

            留言簿(3)

            隨筆分類

            隨筆檔案

            文章分類

            文章檔案

            友情鏈接

            搜索

            •  

            最新評論

            閱讀排行榜

            評論排行榜

            午夜视频久久久久一区 | 精品久久久久久99人妻| 精品久久一区二区三区| 色偷偷888欧美精品久久久| 国产精品久久久天天影视香蕉| 久久精品中文字幕第23页| 午夜视频久久久久一区| 久久精品国产亚洲av水果派 | 精品国产乱码久久久久久1区2区| 国产精品久久亚洲不卡动漫| 久久久久无码中| 久久天天躁狠狠躁夜夜躁2O2O| 国内精品久久久久影院免费| 精品水蜜桃久久久久久久| AV无码久久久久不卡蜜桃| 久久综合久久综合九色| 久久精品国产亚洲AV忘忧草18| www.久久热| 色偷偷久久一区二区三区| 久久精品亚洲福利| AV狠狠色丁香婷婷综合久久| 日日狠狠久久偷偷色综合96蜜桃| 久久精品中文闷骚内射| 一个色综合久久| 日本精品一区二区久久久| 久久精品无码一区二区三区| 欧美丰满熟妇BBB久久久| 亚洲欧美日韩精品久久亚洲区| 一本久久久久久久| 国产精品久久久久久久久| 亚洲欧美伊人久久综合一区二区| 欧美午夜精品久久久久久浪潮| 国产精品美女久久久久av爽| 99久久久精品| 青青草国产精品久久久久| 99久久久精品| 97久久精品午夜一区二区| aaa级精品久久久国产片| 久久伊人精品青青草原高清| 国产一区二区三区久久精品| 久久精品草草草|