链接: https://vjudge.net/problem/LightOJ-1354 题意: An IP address is a 32 bit address formatted in the following way a.b.c.d where a, b, c, d are integers each ranging from 0 to 255. Now you are given two IP addresses, first one in decimal form and secon…

1354 - IP Checking   PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Limit: 32 MB An IP address is a 32 bit address formatted in the following way a.b.c.d where a, b, c, d are integers each ranging from 0 to 255. Now you are given two I…

1354 - IP Checking   PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Limit: 32 MB An IP address is a 32 bit address formatted in the following way a.b.c.d where a, b, c, d are integers each ranging from 0 to 255. Now you are given two I…

一.Description Suppose you are reading byte streams from any device, representing IP addresses. Your task is to convert a 32 characters long sequence of '1s' and '0s' (bits) to a dotted decimal format. A dotted decimal format for an IP address is form…

题意:给出一个n面的色子,问看到每个面的投掷次数期望是多少. 析:这个题很水啊,就是他解释样例解释的太...我鄙视他,,,,, dp[i] 表示 已经看到 i 面的期望是多少,然后两种选择一种是看到新的一面,另一种是看到旧的一面,然后就很出答案了. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include &…

lightoj 1020 A Childhood Game 链接:http://lightoj.com/volume_showproblem.php?problem=1020 题意:一堆石子有 m 个,Alice和Bob可以从中取1个或者2个,如果Alice先取,最后取的人败:如果Bob先取,最后取的人胜.问胜利者. 思路:枚举前几个石子的情况,很容易找出每种情况的必败点,找规律.如果Alice先取,Alice的必败点为mod 3 == 1,如果Bob先取,Bob必败点在 mod 3 == 0处…

http://www.lightoj.com/volume_showproblem.php?problem=1023 题意:26字母全排列 思路:用next_permutation或者思维想一下都可以 /** @Date : 2016-12-17-13.28 * @Author : Lweleth (SoungEarlf@gmail.com) * @Link : https://github.com/ * @Version : */ #include<bits/stdc++.h> #defin…

#include <iostream> #include <cstdio> #include <cmath> using namespace std; const int maxN = 100005; int a[maxN]; int t, tt; int n, q, x, y; int Bsearch_lower_bound(int x) { int l = 0, r = n - 1, mid = 0; while (l <= r) { mid = (l + r…

java有大数模板 import java.util.Scanner; import java.math.*; public class Main { public static void main(String[] args) { Scanner cin =new Scanner(System.in); int T=cin.nextInt(); BigInteger c=BigInteger.valueOf(0); for(int i=1;i<=T;++i){ BigInteger a=cin…

暴力就行了,找出素因子,正的最多是30,然后负的最多是31(这一点wa了一次) #include <cstdio> #include <iostream> #include <ctime> #include <vector> #include <cmath> #include <map> #include <queue> #include <algorithm> #include <cstring>…