2011华为编程大赛第二道题

// OutFunc.cpp : 定义控制台应用程序的入口点。///*2011年华为编程大赛（a卷）2、出圈问题（30分）问题描述M个人围成一圈报数，数到N（1<N<10）的倍数或包含N这个数字时出圈，问剩下的最后一个人在原来的位置是多少？报数规则：1、从第一个人开始报数为1，下一个人报数为上一个人报数加12、报数的最大值为2000，如果报数超过2000，则下一个人重新从1开始报数要求实现函数int OutFunc (unsigned int iTotalNum, unsigned int iKey)【输入】iTotalNum：开始报数前的总人数, 0<iTotalNum<65535iKey：     题目中要求的数目N【输出】无【返回】剩下的人的原来的位置示例输入：iTotalNum =5,  iKey =3返回：4输入：iTotalNum =15,  iKey =3返回：10*/#include "stdafx.h"#include "iostream"using namespace std;int OutFunc (unsigned int iTotalNum, unsigned int iKey){if (iTotalNum == 0 || iKey == 0){return 0;}int *TotalNum = new int[iTotalNum + 1];for (int i = 0; i < iTotalNum; i++){TotalNum[i] = i + 1;}TotalNum[iTotalNum] = '\0';//int *test = TotalNum;int iLeftNum = iTotalNum;int count = 1;int locate = 0;while(true){if(count % iKey == 0 || count % 10 == iKey || (int)(count/10) %10 == iKey || (int)(count/100) % 10 == iKey || (int)(count/1000) * 10 ==iKey){TotalNum[locate] = 0;iLeftNum--;}locate++;if (locate == iTotalNum){locate = 0;}while(TotalNum[locate] == 0){locate++;if (locate == iTotalNum){locate = 0;}}count++;if (count > 2000){count = 1;}if (iLeftNum == 1){break;}}for (int i = 0; i < iTotalNum; i++){if (TotalNum[i] != 0){int output = TotalNum[i];delete [] TotalNum;TotalNum = NULL;return output;}}return 0;}int _tmain(int argc, _TCHAR* argv[]){int iTotalNum, iKey;cout << "请输入总人数：";cin >> iTotalNum;cout << "请输入关键数：";cin >> iKey;cout << "最后剩下的数为：" << OutFunc(iTotalNum,iKey);system("pause");return 0;}

思路比较简单，写出来时候各种临界条件如果真是比赛，时间大大滴不够

思路：设置与人数同大小的整型数组，其内容就是其编号。每报一个数，只要判断刚好是iKey的倍数或包含iKey，则将数组中当前报数的位置置为0。下次报数时候则跳过数值为0的位置。