• <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!!!!?。。。?br>            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(264) 評論(0)  編輯 收藏 引用
            <2011年11月>
            303112345
            6789101112
            13141516171819
            20212223242526
            27282930123
            45678910

            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

            久久亚洲国产欧洲精品一| 国产99久久久久久免费看 | 久久这里有精品视频| 香蕉久久永久视频| AAA级久久久精品无码片| 日韩AV毛片精品久久久| 久久久久久人妻无码| 久久精品无码一区二区三区日韩| 伊人久久久AV老熟妇色| 国产精品日韩深夜福利久久| 久久久久久久久久久精品尤物| 麻豆精品久久久一区二区| 日韩精品久久久久久免费| 久久天堂电影网| 久久丫忘忧草产品| 日韩中文久久| 国产精品热久久无码av| 成人久久久观看免费毛片| 久久精品国产免费观看| 久久亚洲AV永久无码精品| 国产精品女同一区二区久久| 69久久夜色精品国产69| 久久综合给久久狠狠97色 | 久久婷婷国产麻豆91天堂| 久久精品亚洲精品国产色婷| 久久AV高潮AV无码AV| 亚洲日韩欧美一区久久久久我| 久久久久四虎国产精品| 久久国产精品久久精品国产| 国产精品99久久精品| 国产精品久久久久aaaa| 97超级碰碰碰久久久久| 国产欧美久久久精品| 色欲久久久天天天综合网| 日产精品99久久久久久| 青青草原精品99久久精品66| 精品永久久福利一区二区| 国内精品伊人久久久久av一坑| 久久精品蜜芽亚洲国产AV| 久久九九亚洲精品| 国产精品亚洲综合专区片高清久久久 |