http://poj.org/problem?id=1028 题意(水):做一个游览器的历史记录. back:后退到上一个页面,当上一个页面没有时,输出ignored. forward:向前一个页面,但此页面为最前的页面时,输出ignored. vista:游览所指定的页面. quit:退出. 解题思路:题目是说用栈,但不用栈也是可以的.就用纯数组来模拟就栈. 我用的是string类型,这个类型有个好处,就是比较不用strcmp函数,还有输入方便,直接cin就行.但其实本质就是char[]; #…

题目链接:http://poj.org/problem? id=1028 Description Standard web browsers contain features to move backward and forward among the pages recently visited. One way to implement these features is to use two stacks to keep track of the pages that can be rea…

Web Navigation Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 31088   Accepted: 13933 Description Standard web browsers contain features to move backward and forward among the pages recently visited. One way to implement these features…

题目地址:http://poj.org/problem?id=1028 测试样例: Sample Input VISIT http://acm.ashland.edu/ VISIT http://acm.baylor.edu/acmicpc/ BACK BACK BACK FORWARD VISIT http://www.ibm.com/ BACK BACK FORWARD FORWARD FORWARD QUIT Sample Output http://acm.ashland.edu/ ht…

#include <iostream>#include <cstdio>#include <cmath>#include <stack>#include <string> using namespace std; int main(){ char command[16]; char url[71]; stack<string> forwardStack; stack<string> backStack; string cu…

Description Standard web browsers contain features to move backward and forward among the pages recently visited. One way to implement these features is to use two stacks to keep track of the pages that can be reached by moving backward and forward.…

考查代码能力的题目.也能够说是算法水题,呵呵. 推荐新手练习代码能力. 要添加难度就使用纯C实现一下stack,那么就有点难度了,能够使用数组模拟环形栈.做多了,我就直接使用STL了. #include <stdio.h> #include <iostream> #include <stack> #include <string> using namespace std; int main() { stack<string> forward; s…

#include<stdio.h> #include<string.h> int main() { ][]; ]; int i,depth; strcpy(s[],"http://www.acm.org/"); i=; depth=; ) { scanf("%s",str); ) { break; } ) { scanf("%s",&s[++i]); printf("%s\n",s[i]); d…

Web Navigation Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 30828   Accepted: 13821 Description Standard web browsers contain features to move backward and forward among the pages recently visited. One way to implement these features…

注:fjutoj基本每周都有一次周赛,欢迎大家都来参加! 网址:http://59.77.139.92/index.jsp A题:来源 POJ 2773 题意:给两个数m和k,问第k 个和m 互素的数是多少(从1到无穷大). 思路: 二分 + 容斥 先求出m 的素因子p[],数x 和m 互素就意味着x 不存在p 数组中的任意一个素因子,现在要 求n 下面不存在p[]素因子的数的数量可以转化为,n-存在p[]中任意一个素因子的数的个数(经典题型,用容斥可以求),现在二分(k,INF)可以求出答案.…