n 个硬币中有1枚是假币,真假币唯一的区别是假币重量轻,如何快速找出假币
/**
* @description: n 个硬币中有1枚是假币,假币重量轻,如何快速找出假币
* @author: michael ming
* @date: 2019/7/6 20:37
* @modified by:
*/
#include <iostream>
#include <ctime>
#include <random>
using namespace std;
int findcoin(int *weight, int left, int right, int &weightimes)
{
if(left+1 == right)//只有2枚硬币
{
weightimes++;//称重比较一次
if(weight[left] < weight[right])
return left;//返回重量小的位置
else
return right;
}
int i, mid, weightsumL, weightsumR;
weightsumL = weightsumR = 0;
mid = left + (right-left)/2;
if((right-left+1)%2 == 0)//偶数枚银币
{
weightimes++;
for(i = left; i <= mid; ++i)
weightsumL += weight[i];//计算左边重量(计算机没有天平,只能一个个加)
for(i = mid+1; i <= right; ++i)
weightsumR += weight[i];//右边重量
if(weightsumL > weightsumR)//左边重,假币在右边
return findcoin(weight,mid+1,right,weightimes);
else if(weightsumL < weightsumR)//假币在左边
return findcoin(weight,left,mid,weightimes);
else//假币不在两边(偶数枚银币)
;//什么都不做,不必再找了
}
else//奇数枚硬币
{
weightimes++;
for(i = left; i <= mid-1; ++i)
weightsumL += weight[i];//计算左边重量
for(i = mid+1; i <= right; ++i)
weightsumR += weight[i];//右边重量
if(weightsumL > weightsumR)//左边重,假币在右边
return findcoin(weight,mid+1,right,weightimes);
else if(weightsumL < weightsumR)//假币在左边
return findcoin(weight,left,mid-1,weightimes);
else//两边相等(奇数枚硬币),剩余的那个是假币
return mid;
}
}
int main()
{
srand(unsigned(time(0)));
int num, i, weightimes = 0;
cout << "请输入硬币总个数:";
cin >> num;
const int coinNum = num;
int *weight = new int [coinNum];
for(i = 0; i < coinNum; ++i)
{
weight[i] = 10;
}
i = rand()%num;
weight[i] = 9; //随机生成假币
for(i = 0; i < coinNum; ++i)//打印硬币信息
{
cout << i + 1 << " 硬币重量: " << weight[i] << endl;
}
cout << "假硬币是第" << findcoin(weight,0,coinNum-1,weightimes)+1 << "个。" << endl;
cout << "共称了" << weightimes << "次,找到假币。";
delete[]weight;
return 0;
}
输入 2500枚、5001枚,100万枚,最多需要 log2n 向上取整次就能找到。