算法设计 分析 伪币问题

前提
int FalseCoin(int a[], int left, int right)
//设 left,right为a[]逻辑地址
{ 
  int mid;
 int temp=0;
  if((left+right+1)%2)
     {
    temp=right;
     right=right-1;
     mid = (left + right+1) / 2;
       }
 else
   min=(left + right+1) / 2;
 int sum1 = 0, sum2 = 0;
 int i;

    

    if ((right-left) == 1)//只有两枚硬币，轻者为伪币，等重则无伪币  

        return (a[left] < a[right]) ? left : (a[left] > a[right]) ? right : -1;

 

    //多于两枚硬币，将数组分成两半，分而治之 

    for (i=left; i<=mid; i++)

        sum1 += a[i];