继续发！把Timer2做成72M的捕获，

以9600波特率发送数据，捕获存储，得到数组ary！
void Auto_baud(void)
{        u16 i;                         
        const u16 tmplen=100;//捕获个数
        u16 ary[tmplen];         //捕获数组

        TIM_TimeBaseInitTypeDef          TIM_TimeBaseStructure;       

//Timer2--72M: 时基减得cap
         TIM_DeInit(TIM2);//复位TIM2定时器                 
         TIM_TimeBaseStructure.TIM_Period = 65535;                 //计数值max65535     
         TIM_TimeBaseStructure.TIM_Prescaler = 0;                //NO分频72M   
         TIM_TimeBaseStructure.TIM_ClockDivision = 0x0; // 时钟分割  
         TIM_TimeBaseStructure.TIM_CounterMode = TIM_CounterMode_Up;  //计数方向向上计数
         TIM_TimeBaseInit(TIM2, &TIM_TimeBaseStructure);

         TIM_Cmd(TIM2, ENABLE);        //开启时钟       

////////////
         while(U3RX==1)TIM2->CNT = 0;          //等第一个下降沿
         while(U3RX==0)TIM2->CNT = 0;          //等第一个下降沿
         while(U3RX==1)TIM2->CNT = 0;          //等第一个下降沿
         for(i=0;i<tmplen;i++){

                 while(U3RX==0);
               
                 ary = TIM2->CNT;
                 
                  while(U3RX==1)TIM2->CNT = 0;
                 }

         //发回数组
                 for(i=0;i<tmplen;i++){
                 printf("%d,",ary);
                 }



数组里的数据如下：
984,14998,7702,22662,7686,37638,7702,7702,14982,14982,7686,7702,7686,7702,14998,7702,7686,29958,7686,22678,7318,14982,7702,22662,14982,7686,14982,14982,37638,14982,29958,14982,7702,7702,7686,14982,7702,7702,7702,14998,14982,7302,14982,37654,14998,37638,14998,37638,14982,22662,7302,14982,7686,37638,7686,22678,7318,14982,14982,14982,14982,29958,14982,7702,7702,7686,14982,7702,7686,14982,14982,7302,14982,22662,7686,14982,7686,22662,14998,22662,7686,14998,7686,7702,14998,22662,7302,14982,14998,14998,7702,22662,7302,14982,7702,22678,14998,7686,22662,14998,


9600波特率，对于Timer2的捕获值理论是7494.543972
可以看出，基本上捕获的都是这个的整数倍~~~~~

现在问题是，如何根据捕获数组求得最小公约数值~~