一、题目描述Let us define a regular brackets sequence in the following way: Empty sequence is a regular sequence. If S is a regular sequence, then (S) , [S] and {S} are both regular sequences. ...
一、题目描述Let us define a regular brackets sequence in the following way: Empty sequence is a regular sequence. If S is a regular sequence, then (S) , [S] and {S} are both regular sequences. ...
1198. Substring Constraints Time Limit: 1 secs, Memory Limit: 32 MB Description Dr lee cuts a string S into N pieces,s[1],…,s[N]....Now, Dr lee gives you these N sub-strings: s[1],…s[N]....
编号为1-n的火车经过一个栈后,判断能否实现输入的序列。 思想是将输入元素以数组形式存储,将1-n的序列放入栈中,然后将栈顶元素与数组中的首元素进行比对,若相同则将栈顶元素pop掉,并将数组元素向后加一,这样...
1129. ISBN Total: 4941 Accepted: 1697 Rating: 2.0/5.0(15 votes) 012 345 Time Limit: 10sec Memory Limit:
标签: sicily
水题,不过个别测试用例还需要注意,开始由于忘记了0而WA了2次。 题目的主要思想是先判断字符中每一位是否为数字,若是数字,则进行题目中要求的运算。最后求和,求余。 代码如下: #include ...
sicily1031Campus之最短路径解题报告题目陷阱昨天讲了人工智能中的搜索算法,贪婪算法,A*算法,爬山算法,模拟退火算法和遗传算法,听着好屌的样子,其实也没什么,回来之后打算练习一个A*算法的实例,刚想做A*算法...
完全背包 #include #include int dp[320]; int n,c; int coins[10]; int main() { int i,j; while (scanf("%d",&n)!=EOF) { scanf("%d",&c); for (i=0;i;i++) scanf("%d
题目:1029. Rabbit 题意: • 开始有一对成年兔子 • 每对成年兔子每个月产生一对小兔子 • 每只小兔子经过m个月变成成年兔子 • 问经过d个月后有多少兔子 • 约束: 1 , 1 解法:递推 ...
1139. 电路稳定性 Constraints Time Limit: 1 secs, Memory Limit: 32 MB Description ...zzz有一个电路,电路上有n个元件。已知元件i损坏而断开的概率是Pi(i=1,2,...,n,0≤Pi≤1)。...1. 一个元件是最小
11157. Crossword Time Limit: 1sec Memory Limit:256MB ...Mirko has assembled an excellent crossword puzzle and now he wants to frame it....
2302. Queue Implementation Using a Circular Array Constraints Time Limit: 1 secs, Memory Limit: 256 MB , Framework Judge Description template class Queue { public: ...
Description 使用线性探测法(LinearProbing)可以解决哈希中的冲突问题,其基本思想是:设哈希函数为h(key)=d,并且假定哈希的存储结构是循环数组,则当冲突发生时,继续探测d+1,d+2…,直到冲突得到解决. ...
1034. Forest 限制条件 时间限制: 1 秒, 内存限制: 32 兆 ... In the field of computer science, forest is important and deeply researched , it is a model for many data structures ....
最大流经典问题
因为今天中午,不知道为何,sicily服务器挂了,
一、题目描述Given a sequence A = , a2, …, am >, let sequence B = , b2, …, bk > be a subsequence of A if there exists a strictly increasing sequence ( i1 …, ik ) of indices of A such that fo
1443. Printer Queue Constraints Time Limit: 1 secs, Memory Limit: 32 MB Description The only printer in the computer science students' union is experiencing an extremely heavy workload....
1048. Inverso Constraints Time Limit: 1 secs, Memory Limit: 32 MB Description The game of ‘Inverso’ is played on a 3x3 grid of colored fields (each field is either black or white)....
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <vector> #include <map> #include <......
求解指定周长的直角三角形的解
sicily 1153 马的周游问题。棋盘范围是8*8
直接偷懒调用了sort哈哈哈哈哈哈哈哈 1 #include<iostream> 2 #include<algorithm> 3 using namespace std; 4 int main() { 5 int T; 6 cin >> T; ... 8 int a...
1148. 过河 Constraints Time Limit: 1 secs, Memory Limit: 32 MB Description 在河上有一座独木桥,一只青蛙想沿着独木桥从河的一侧跳到另一侧。在桥上有一些石子,青蛙很讨厌踩在这些石子上。...
一. 模仿树的的先序遍历。范围是1000个节点。用数组存储节点的信息。 二. 要注意的是,头结点是不确定的,所以在前序遍历之前要找出头结点,除了头结点的下标值出现一次之外,其他结点... 3 // sicily-1156 4 /...
Sicily.1001.Alphacode 一道经典的DP题。一看题就去百度找答案了,找到不少博文是解这道题,但都只是把解决方法写出来,没有解释,那我来解释一下: 首先,解这道题的基本思路(不考虑各种极端情况)是: 一个...
由于中大的oj需要内网才能进去,就提供不了原始题目了,但是题目的意思就是说,开始有一对成年兔子,一对成年兔子每年能生一对幼兔,幼兔等m个月才成长为成年兔子,问d个月后总共有多少对兔子。...
一. 题意 按照孩子们需要的积木块数排序(从小到大),先处理需要积木块数少的孩子。... 3 // sicily-1134 4 // 5 // Created by ashley on 14-10-25. 6 // Copyright (c) 2014年 ashley. All rights r...
1910. Box Constraints Time Limit: 5 secs, Memory Limit: 32 MB Description There are N boxes on the ground, which are labeled by numbers from 1 to N . The boxes are magical, the ...
1913. Slides Constraints Time Limit: 1 secs, Memory Limit: 32 MB Description There are N slides lying on the table. Each of them is transparent and formed as a rectangle....
1510. Mispelling Constraints Time Limit: 1 secs, Memory Limit: 32 MB Description Misspelling is an art form that students seem to excel at. Write a program that removes the n th character ...