学步园

问题描述

十个人围成一圈，分别编号一到十，从一开始报数报到七的出去，继续报，问最后剩下的是原来的几号？

代码1(本问题解决方案）
#include <stdio.h><br /> #include <stdlib.h><br /> #define LENGTH 10 //十个人<br /> #define KEY 7 //被淘汰的编号<br /> /*<br /> * 思路：<br /> * 用一个数组存放10个人的状态。1存货，0淘汰。<br /> * 当只有一个人剩下，则结束。<br /> */<br /> int<br /> main(void)<br /> {<br /> int people[LENGTH];<br /> int alive = LENGTH;//没有淘汰的人数<br /> int no;//编号<br /> int i;</p> <p> for (i = 0; i < LENGTH; ++i) //状态全部置1<br /> people[i] = 1;<br /> for(no = 1,i = 1; alive !=1; ++i) {//只有剩下一个人则停止循环<br /> if (people[i%(LENGTH + 1) - 1]) {<br /> if (no%KEY == 0) {//淘汰该人<br /> people[i%(LENGTH + 1) -1] =0;<br /> alive--;<br /> }<br /> no++;<br /> }<br /> }<br /> for (i = 0; people[i] != 1; i++) //找出存货的人<br /> ;<br /> printf("第%d个人存活/n", i+1);</p> <p> exit(0);<br /> }

代码2（代参数的灵活方案）
#include <stdio.h><br /> #include <stdlib.h><br /> #include <assert.h></p> <p>int<br /> main(int argc, char *argv[])<br /> {<br /> int *people;<br /> int people_counts;<br /> int key_num;<br /> int alive ;//没有淘汰的人数<br /> int no;//编号<br /> int i;</p> <p> assert(argc == 3); //验证参数<br /> assert(people_counts = atoi(argv[1]));<br /> assert(key_num = atoi(argv[2]));<br /> assert(people = (int *)malloc(people_counts * sizeof (int)));//动态分配<br /> for (i = 0; i < people_counts; ++i) //状态全部置1<br /> people[i] = 1;<br /> for(no = 1,i = 1, alive = people_counts;<br /> alive !=1; ++i) {//只有剩下一个人则停止循环<br /> if (people[i%(people_counts + 1) - 1]) {<br /> if (no%key_num == 0) {//淘汰该人<br /> people[i%(people_counts + 1) -1] =0;<br /> alive--;<br /> }<br /> no++;<br /> }<br /> }<br /> for (i = 0; people[i] != 1; i++) //找出存货的人<br /> ;<br /> printf("第%d个人存活/n", i+1);<br /> if (people)//释放<br /> free(people);<br /> exit(EXIT_SUCCESS);<br /> }