数学趣题——寻找假币

递归分治二分查找法

#include "stdio.h"

int getFalseCoin(int coin[], int low, int high)
{
    int i, sum1 = 0, sum2 = 0, sum3 = 0;

    if(low + 1 == high)
    {
        if(coin[low] < coin[high]) return low + 1;
        else return high + 1;
    }

    if((high - low + 1) % 2 == 0)                    /*n是偶数*/
    {
        for(i = low; i <= low + (high - low) / 2; i++)
            sum1 = sum1 + coin[i];               /*前半段和*/

        for(i = low + (high - low) / 2 + 1; i <= high; i++)
            sum2 = sum2 + coin[i];               /*后半段和*/

        if(sum1 < sum2) return getFalseCoin(coin, low, low + (high - low) / 2);

        if(sum1 > sum2) return getFalseCoin(coin, low + (high - low) / 2 + 1, high);
    }

    if((high - low + 1) % 2 != 0)
    {
        for(i = low; i <= low + (high - low) / 2 - 1; i++)
            sum1 = sum1 + coin[i];               /*前半段和*/

        for(i = low + (high - low) / 2 + 1; i <= high; i++)
            sum2 = sum2 + coin[i];               /*后半段和*/

        sum3 = coin[low+(high-low)/2];

        if(sum1 < sum2)
            return getFalseCoin(coin, low, low + (high - low) / 2 - 1);

        if(sum1 > sum2)
            return getFalseCoin(coin, low + (high - low) / 2 + 1, high);

        if(sum1 + sum3 == sum2 + sum3) return low + (high - low) / 2 + 1;
    }

}

int main()
{
    int coin[30] = {2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2};
    printf("The %dth coin is false\n", getFalseCoin(coin, 0, 29));
    return 0;
}