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1
2、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<65535
iKey:
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 <cstdio> #include <vector> #include <algorithm> using namespace std; typedef struct PERSON { PERSON(){}; PERSON(unsigned int pos,unsigned int num):pos(pos),num(num){}; unsigned int pos; unsigned int num; }PSN; int OutFunc (unsigned int iTotalNum, unsigned int iKey) { vector<PSN> a; vector<PSN>::iterator it; for(unsigned int i = 0;i<iTotalNum;i++) { PSN p(i+1,((i+1)%2000)?((i+1)%2000):2000); a.push_back(p); } //printf("%d\n",a.size()); int tmp; while(a.size() > 1) { tmp = a.size(); for(it = a.begin();it!=a.end();it++) { if((it->num%iKey == 0) || (it->num%10 == iKey) || ((it->num/10)%10) == iKey ||(it->num/100)%10 == iKey || (it->num/1000)%10 == iKey) { if(it != a.begin()) { a.erase(it); it--; } else { a.erase(it); it = &a[tmp-2]; } } else { if(it == a.begin()) it->num = a.at(tmp-1).num+1; else { it--; unsigned int b = it->num; (++it)->num = ((b+1)%2000)?((b+1)%2000):2000; } } } } return a.begin()->pos; } int main() { unsigned int iTotalNum = 5; unsigned int iKey = 3; printf("%u\n",OutFunc(iTotalNum,iKey)); return 0; }