解决下约瑟夫环问题

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>#include <stdlib.h>#define LENGTH 10 //十个人#define KEY 7 //被淘汰的编号/* * 思路： * 用一个数组存放10个人的状态。1存货，0淘汰。 * 当只有一个人剩下，则结束。 */intmain(void){intpeople[LENGTH];intalive = LENGTH;//没有淘汰的人数intno;//编号inti;for (i = 0; i < LENGTH; ++i) //状态全部置1people[i] = 1;for(no = 1,i = 1; alive !=1; ++i) {//只有剩下一个人则停止循环if (people[i%(LENGTH + 1) - 1]) {if (no%KEY == 0) {//淘汰该人people[i%(LENGTH + 1) -1] =0;alive--;}no++;}}for (i = 0; people[i] != 1; i++) //找出存货的人;printf("第%d个人存活/n", i+1);exit(0);}
代码2（代参数的灵活方案）
#include <stdio.h>#include <stdlib.h>#include <assert.h>intmain(int argc, char *argv[]){int*people;intpeople_counts;intkey_num;intalive ;//没有淘汰的人数intno;//编号inti;assert(argc == 3); //验证参数assert(people_counts = atoi(argv[1]));assert(key_num = atoi(argv[2]));assert(people = (int *)malloc(people_counts * sizeof (int)));//动态分配for (i = 0; i < people_counts; ++i) //状态全部置1people[i] = 1;for(no = 1,i = 1, alive = people_counts; alive !=1; ++i) {//只有剩下一个人则停止循环if (people[i%(people_counts + 1) - 1]) {if (no%key_num == 0) {//淘汰该人people[i%(people_counts + 1) -1] =0;alive--;}no++;}}for (i = 0; people[i] != 1; i++) //找出存货的人;printf("第%d个人存活/n", i+1);if (people)//释放free(people);exit(EXIT_SUCCESS);}