Find bad coins

Problem Description

Among n (n>=3) coins, there is an unqualified coin ( too light or too heavy ). If there is only one balance that can be used for weighing and there is no limit to the number of coins to be weighed, design an algorithm and find Take out this unqualified coin, making the number of weighings the least (better than O(n)).

(Hint: Divide into n/3 or n/4, at least two of the same quantity)

Ideas

Drawing on the idea of ​​dichotomy, it is just divided into three piles, two of which have the same number, and the other one can be inconsistent. Regardless of whether it is an odd or even number of coins, (divided by 3 is best) can be divided into 3 roughly the same piles. Compare the two piles if they are the same. If they are not equal, the bad coins are in the two piles. If they are equal, they are in the third pile. Repeat the operation until there are only 2 coins left.

 

//Q6   在n(n>=3)枚硬币中有一枚重量不合格的硬币(过轻或过重)，如果只有一架天平可以用来称重且称重的硬币数没有限制，
// 设计一个算法，找出这枚不合格的硬币，使得称重的次数最少(优于O(n))。

// 思路： 采用分治策略。如果n<3，那么将拿走的硬币与剩下的硬币比较，不等的是坏币；将n枚硬币分为大致相等的3份，
//如果n(mod 3)≠0，那么令两份少的硬币相等。取两份相等的硬币放到天平上，如果不等，那么这两份硬币中含有坏币，
//否则坏币在第3份中。递归处理，直到n<3为止。

//关键词：硬币 坏 分治coins

//硬币结构体
struct Coin
{
    int weight;
    int label;
    Coin(int weight, int label)// 构造函数
    {
        this->weight = weight;
        this->label = label;
    }
};

int gettotalweight(vector<Coin> coins) {
    int totalweight = 0;
    for (vector<Coin>::iterator iter = coins.begin(); iter != coins.end(); ++iter) {
        totalweight = totalweight+iter->weight;
    }
    return totalweight;
}

#include "algorithm"
int turewight = 0;
int findBadCoin(vector<Coin> coins) {
    int n = coins.size();
    int k = n / 3; // 将硬币分为3堆 个数分别为k,k,n-2k
    vector<Coin> A,B,C;  
    int result = 0;
    A.insert(A.begin(),coins.begin() , coins.begin() + k);
    B.insert(B.begin(),coins.begin() + k, coins.begin() + 2*k );
    C.insert(C.begin(),coins.begin() + 2*k, coins.end());
    if (n == 2)
    {
        coins[0].weight == turewight ? result = coins[1].label : result = coins[0].label;
        return result;
    }
    if(n>3){
        if (gettotalweight(A)!= gettotalweight(B)) { //A与B的重量不相等
            coins.clear();
            coins.insert(coins.begin(), A.begin(), A.end()); // 重新整合A和B的元素到coins中
            coins.insert(coins.end(), B.begin(), B.end());
            turewight = C[0].weight;//记录真实重量方便最后的比较
        }
        else // 坏币在第3堆中
        {
            coins.clear();
            coins.insert(coins.begin(), C.begin(), C.end());
        }      
        return findBadCoin(coins);
    }  
}
//Ques20201004
int main()
{
    //Q1
    //cin >> k;//输入要找的数字c语言把cin换为scanf即可
    //cout << BinarySearch(0, 9);//从数组a[0]到a[9]c语言把cout换为printf即可

    //Q2
    /*int n = 9;
    int arr[] = { -1,3,5,2,7,8,-9,1,-4 };
    
    f(arr,n);*/


    // Q3
 /*   char A = 'A';
    char B = 'B';
    char C = 'C';
    Hanoi(3, A, B, C);*/



    //Q4
   /* int arr[] = { 1,3,5,8,9,7,6,1,2 };
    cout<<siglePeak(arr,0,8);*/


    // Q5
    //int arr[] = { 1,1,2,3,5,7,8,9 };
    //printL_U(arr,1,7,8);// L=1，U=7 数组长度为8

    // Q6
    vector<Coin> coins;
    for (int i = 0; i < 19;i++) {
        
        if (i == 12) {// 坏硬币的重量为2           
            coins.push_back(Coin(2, 12));
        }
        else { // 生成20个硬币
            coins.push_back(Coin(1, i));
        }
    }
    cout<<findBadCoin(coins);
    return 0;
}