• <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>
            voip
            風的方向
            厚德致遠,博學敦行!
            posts - 52,comments - 21,trackbacks - 0
                           有時候失去的總比得到的多!

            Description

            The Double Seventh Festival, on the 7th day of the 7th lunar month, is a traditional festival full of romance. On that day, the God of the Love came to the digital kingdom, and annunciated to the people:

            Notification


            Do you want to know who is your fere? You can find him or her by this means:
            Please add all the factors of your ID-Card-Number, and you can find this is your fere's
            ID-Card-Number.


            The factors of a numer N is the numbers,they being small than the number N ,and can being divided exactly by the number N.
            For example, the number 12 has the factors, such as 1,2,3,4,6.

             

            Input

            The fist line is the number T (1 <= T <= 500000), indicated the number of the test cases.
            The next T lines, each line has one integer N( 1 <= N <= 500000), meaning the ID-Card-Number.

            Output

            For each test case, you should output the fere's ID-Card-Number on a line.

            Sample Input

            3
            2
            10
            20
            

             

            Sample Output

            1
            8
            22
                  這個題目在我剛開始接觸ACM的時候做出來最有成就的一個!求各因子之和!
            代碼如下:
            #include<cstdio>
            int main()
            {
                
            int i,j,n,m,sum;
                scanf(
            "%d",&n);
                
            for(i=0;i<n;i++)
                
            {
                    sum
            =0;
                    scanf(
            "%d",&m);
                    
            for(j=1;j*j<=m;j++)
                    
            {
                        
            if(m%j==0)
                        
            {
                            sum
            =sum+j+m/j;
                        }

                        
            if(j*j==m)
                            sum
            -=j;
                    }

                    printf(
            "%d\n",sum-m);
                }

                
            return 0;
            }

            posted on 2010-09-06 09:34 jince 閱讀(1058) 評論(0)  編輯 收藏 引用 所屬分類: Questions
            哈哈哈哈哈哈
            久久久久久久免费视频| 一级做a爰片久久毛片人呢| 亚洲国产精品一区二区久久hs| 亚洲综合精品香蕉久久网| 久久丫精品国产亚洲av| 情人伊人久久综合亚洲| 亚洲国产精品嫩草影院久久| 精品永久久福利一区二区| 2021国内久久精品| 伊人色综合久久天天| 一本一道久久a久久精品综合| 91久久精品91久久性色| 精品无码久久久久国产动漫3d| 久久综合丝袜日本网| 一本色道久久综合亚洲精品| 久久精品国产亚洲AV不卡| 久久国产亚洲精品无码| 久久精品国产免费观看| 日本精品一区二区久久久| 久久精品成人免费看| 久久无码人妻一区二区三区| 久久无码中文字幕东京热| 人妻少妇精品久久| 国产激情久久久久影院小草| 久久91精品国产91久久小草| 日韩精品久久久久久免费| 欧美精品乱码99久久蜜桃| 国产欧美久久久精品影院| 久久青青草原精品国产软件| 久久se这里只有精品| 久久亚洲精品中文字幕三区| 久久综合综合久久97色| 77777亚洲午夜久久多喷| 99久久这里只有精品| 久久99精品国产99久久6男男| 久久久久亚洲AV无码专区体验| 久久综合色老色| 欧美黑人又粗又大久久久| 国产∨亚洲V天堂无码久久久| 国产精品久久久久aaaa| 欧美久久精品一级c片片|