hdu4727 The Number Off of FFF】的更多相关文章

理解错题意,wa了几次. 我一开始的理解忽略了实际背景,认为错报是绝对的,不依赖于其左边的人. 而实际上某士兵报数的对错取决且仅取决于他所报的数与其左邻所报的数. 所以假设第一个人没有报错,则其后必有人报错,报错时满足s[i] != s[i - 1] + 1 否则第一个人报错. acm.hdu.edu.cn/showproblem.php?pid=4727   #include <cstdio> using namespace std; ; int n, ans; int s[maxn]; i…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4727 水题.. //STATUS:C++_AC_187MS_288KB #include <functional> #include <algorithm> #include <iostream> //#include <ext/rope> #include <fstream> #include <sstream> #include &l…
题目传送门 /* 水题:判断前后的差值是否为1,b[i]记录差值,若没有找到,则是第一个出错 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <string> #include <cmath> using namespace std; ; const int INF = 0x3f3f3f3f; int a…
The Number Off of FFF Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2813 Accepted Submission(s): 935 Problem Description X soldiers from the famous "FFF army" is standing in a line, from le…
The Number Off of FFF Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 602 Accepted Submission(s): 284 Problem Description X soldiers from the famous " *FFF* army" is standing in a line, from…
The Number Off of FFF Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 78    Accepted Submission(s): 36 Problem Description X soldiers from the famous "*FFF* army" is standing in a line, fro…
The Number Off of FFF Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 169    Accepted Submission(s): 83 Problem Description X soldiers from the famous "*FFF* army" is standing in a line, f…
X soldiers from the famous “FFF army'' is standing in a line, from left to right. You, as the captain of FFF, decides to have a number off, that is, each soldier, from left to right, calls out a number. The first soldier should call One, each other s…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4727 题目大意:队列里所有人进行报数,要找出报错的那个人 思路:,只要找出序列中与钱一个人的数字差不是1的人即可,但是要注意的是这些人是从一个队列中间截取下来的,所以很有可能第一个人就报错了,一开始没有考虑这种状况所以出错了 #include<cstdio> #include <iostream> using namespace std; int main() { int t,n,an…
HTML 5 的表单中有着丰富的input种类,比如说input[type="number"],就可以保证用户输入数字,但是input也有一些不好的地方——带有 spinner. 一.spinner 是什么? 简单来说spinner就是数字输入框右边的上下箭头. 二.怎么去除 方法1: /*input number remove spinner*/ input[type="number"]::-webkit-textfield-decoration-containe…