文章目錄

• • 1. 問題描述
  • 2. 解題思路
  • 3. 代碼實現

1. 問題描述

n 個硬幣中有1枚是假幣，真假幣唯一的區別是假幣重量輕，如何快速找出假幣

2. 解題思路

• 暴力做法，一個一個的稱重，O（n）復雜度
• 分治思路

將硬幣等分成兩份，若為奇數，多出一枚，放在天平兩邊

輕的一邊包含假幣，若相等，則假幣是多出的那一枚

對輕的一邊繼續上述操作，直到找出假幣
復雜度O（log n）

3. 代碼實現

/*** @description: n 個硬幣中有1枚是假幣，假幣重量輕，如何快速找出假幣* @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枚硬幣{weightimes++;//稱重比較一次if(weight[left] < weight[right])return left;//返回重量小的位置elsereturn right;}int i, mid, weightsumL, weightsumR;weightsumL = weightsumR = 0;mid = left + (right-left)/2;if((right-left+1)%2 == 0)//偶數枚銀幣{weightimes++;for(i = left; i <= mid; ++i)weightsumL += weight[i];//計算左邊重量(計算機沒有天平，只能一個個加)for(i = mid+1; i <= right; ++i)weightsumR += weight[i];//右邊重量if(weightsumL > weightsumR)//左邊重，假幣在右邊return findcoin(weight,mid+1,right,weightimes);else if(weightsumL < weightsumR)//假幣在左邊return findcoin(weight,left,mid,weightimes);else//假幣不在兩邊（偶數枚銀幣）;//什么都不做，不必再找了}else//奇數枚硬幣{weightimes++;for(i = left; i <= mid-1; ++i)weightsumL += weight[i];//計算左邊重量for(i = mid+1; i <= right; ++i)weightsumR += weight[i];//右邊重量if(weightsumL > weightsumR)//左邊重，假幣在右邊return findcoin(weight,mid+1,right,weightimes);else if(weightsumL < weightsumR)//假幣在左邊return findcoin(weight,left,mid-1,weightimes);else//兩邊相等（奇數枚硬幣），剩余的那個是假幣return mid;} } int main() {srand(unsigned(time(0)));int num, i, weightimes = 0;cout << "請輸入硬幣總個數:";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; //隨機生成假幣for(i = 0; i < coinNum; ++i)//打印硬幣信息{cout << i + 1 << " 硬幣重量: " << weight[i] << endl;}cout << "假硬幣是第" << findcoin(weight,0,coinNum-1,weightimes)+1 << "個。" << endl;cout << "共稱了" << weightimes << "次，找到假幣。";delete[]weight;return 0; }

輸入 2500枚、5001枚，100萬枚，最多需要 log₂n 向上取整次就能找到。

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