Question

我们有两个由字符a-z组成的无限重复消息。每个角色需要不同的时间单位进行传输; a = 1，b = 2，c = 4，d = 8，e = 16 ......，字符|告诉我们当前在消息中的位置。我们的工作是说同步这两条消息需要多少时间。同步意味着两个消息同时开始。

示例：

我们有消息1：ea | babab

和消息2：d | abaca

因此，我们知道消息1是bababea，消息2是：abacad。

消息将以42个单位时间同步：

EA | bababea | = 42

d | abacad | abacad | = 42

示例2： message1：| acabbaaa消息2：| dcbabaaaa。

解决方案：0，因为它们已经同步。

我们想要一个算法来计算第一次同步发生的时间。

我在C中写这个。除了算法本身，我基本上完成了所有工作。

我认为这可以使用扩展的欧几里德算法来完成。

Answer 1

我已回答了不同的question，我认为解决这个问题的方法完全相同。你需要求解方程 m1Offset +（m1Len * intN1）= m2Offset +（m2Len * intN2）

#include <stdio.h>

void findVal(unsigned int m1Offset, unsigned int m1Len, unsigned int m2Offset, unsigned int m2Len) ;
unsigned int getGCD(unsigned int n1, unsigned int n2);

int main()
{
    findVal(17, 25, 8, 17);
    return 0;
}

void findVal(unsigned int m1Offset, unsigned int m1Len, unsigned int m2Offset, unsigned int m2Len) {

    unsigned int  n1       = 0;
    unsigned int  n2       = 0;
    unsigned char foundVal = 1;
    unsigned int  m1Time   = m1Offset;
    unsigned int  m2Time   = m2Offset;

    //No need to find n1 and n2 if msgs are starting from beginning.
    if(((m1Offset == m1Len) && (m2Offset == m2Len)) || ((0 == m1Offset) && (0 == m2Offset)))
    {
        m1Time = 0;
        m2Time = 0;
    }

    //No need to find n1 and n2 if offset times are same.
    else if(m1Offset != m2Offset)
    {
       //Offset times are not same.
       foundVal = 0;

       //Find GCD of m1Len and m2Len.
       unsigned int gcd = getGCD(m1Len, m2Len);

       //There is a solution only if the difference of offsets is divisible by gcd.
       if(0 == (m2Offset-m1Offset) % gcd)
       {
           for(n2=1; n2<(unsigned int)-1; n2++)
           {
               unsigned int temp1 = (m2Len*n2)+(m2Offset-m1Offset);
               if(0 == temp1 % m1Len)
               {
                    n1 = temp1/m1Len;
                    m1Time = m1Offset + n1*m1Len;
                    m2Time = m2Offset + n2*m2Len;

                    foundVal = 1;
                    break;
               }
           }
       }
    }

    if(1 == foundVal)
    {
        printf("Found n1[%u] n2[%u] m1Time[%u] m2Time[%u]\n", n1, n2, m1Time, m2Time);
    }
    else
    {
        printf("Could not find n1, n2, m1Time, m2Time\n");
    }
}

unsigned int getGCD(unsigned int n1, unsigned int n2)
{
    while(n1!=n2)
    {
        if(n1 > n2)
            n1 -= n2;
        else
            n2 -= n1;
    }
    printf("GCD = %u\n",n1);

    return n1;
}

我们需要找到满足上述等式的intN1和intN2。 当且仅当m1Len和m2Len的GCD除以（m2Offset - m1Offset）时才会有解决方案。

在下面的代码中，

m1Len和m2Len：消息长度分别为m1和m2。例如：对于＆＃34; ea | babab＆＃34;长度是＆＃34; bababea＆＃34; = 25的长度，在消息＆＃34; d | abaca＆＃34;长度＆＃34; abacad＆＃34; = 17
m1Offset和m2Offset：初始偏移量。例如：在消息＆＃34; ea | babab＆＃34;，＆＃34; ea＆＃34;是偏移量，它等于17.类似地在＆＃34; d | abaca＆＃34;，＆＃34; d＆＃34;是偏移量，等于8.
m1Time和m2Time应该相等，这是第一个同步时间。

这是我的代码。

Found n1[1] n2[2] m1Time[42] m2Time[42]

输出：

function emailStatusReport() {

  // Set variables. 
  var x = 13;
  var ss = SpreadsheetApp.getActiveSpreadsheet();
  var sheetName = ss.getSheetByName('data');
  var template = HtmlService.createTemplateFromFile("report.html");
  var message;
  var subject;
  var emailQuotaRemaining = MailApp.getRemainingDailyQuota();    
  var theRange = ss.getDataRange();
  var completeMessage = "";
     //Logger.log(theRang.getA1Notation());
  var theManualRange = ss.getRange("A1:C7");
    // var dateTimeStamp = now();
  var vals = theManualRange.getValues();
       for(i=0; i<vals.length; i++){
         var itemNumber = vals[i][0];
         var title = vals[i][1];
         var description = vals[i][2];
         // Logger.log(itemNumber  + " " + title + " " + description + " ");
         completeMessage = completeMessage + ("<BR>" + itemNumber  + " " + title + " " + description + "</BR>");
            }

           Logger.log(completeMessage);

//  Logger.log (x);
//  Logger.log (ss);
//  Logger.log (sheetName);
//  Logger.log ("Remaining email quota: " + emailQuotaRemaining);
//  Logger.log (vals[i][0]);



subject = "Daily Status Report TEST..."
message = "<HTML><BODY>" +          
         "<Font Face='verdana' color='green' size='2'>" + 
         "<P><P><B>Daily Status Report TEST</B>" +
         "<P>Please review the report listed below.  " + 
            // "<P>The value of x is " + x + "." + 

         "<P>" + completeMessage + 
         "<P><P>" + 
         "For the most up-to-date information, please visit: <BR>" + 
         "https://docs.google.com/spreadsheets/d/1jA4XQa7MgQI7OOLwCcmGv7BcUKSjdIPyh6B0CeQI-PI/edit#gid=0" +
         "<P>" + 

         "<P><B>Make it an amazing day!</B>" +

            // "<P>Report generated at: " + dateTimeStamp + ". " +
         "<P>Remaining daily email quota: " + emailQuotaRemaining + ". " +          
             // "<P>This report runs between ___ AM and ___ AM." +

           "</HTML></BODY>"; 


MailApp.sendEmail("greg@gregbartonmba.com", subject, "", {htmlBody: message});


//  Logger.log("Subject is :" + subject); 
//  Logger.log("Message is: " + message);




}

消息同步算法

1 个答案: