题目链接:http://poj.org/problem?id=1686 思路分析:该问题为表达式求值问题,对于字母使用浮点数替换即可,因为输入中的数字只能是单个digit. 代码如下: #include <iostream> #include <cstring> #include <cstdio> #include <cmath> #include <cstdlib> #include <string> using namespace…

Description A math instructor is too lazy to grade a question in the exam papers in which students are supposed to produce a complicated formula for the question asked. Students may write correct answers in different forms which makes grading very ha…

Problem Description A math instructor is too lazy to grade a question in the exam papers in which students are supposed to produce a complicated formula for the question asked. Students may write correct answers in different forms which makes grading…

原题目网址:http://poj.org/problem?id=1686 题目中文翻译: Description 数学教师懒得在考卷中给一个问题评分,因为这个问题中,学生会为所问的问题提出一个复杂的公式,但是学生可以用不同的形式写出正确的答案,这使得评分非常困难. 所以,教师需要计算机程序员的帮助,或许你可以提供帮助. 你应该编写一个程序来阅读不同的公式,并确定它们是否在算术上相同.   Input 输入的第一行包含一个整数N(1 <= N <= 20),即测试用例的数量. 在第一行之后,每个…

  Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 3721   Accepted: 1290 Description A math instructor is too lazy to grade a question in the exam papers in which students are supposed to produce a complicated formula for the question ask…

因为这个题目说明了优先级的规定,所以可以从左到右直接运算,在处理嵌套括号的时候,可以使用递归的方法,给定每一个括号的左右边界,伪代码如下: int Cal(){ if(括号)  sum += Cal(); else sum += num; return sum; } 但是这个题目着实坑了我一下,见过WA了,没见过TLE呢……我因为没有看到有空格这个条件,无线TLE,又是消除函数又是改用数组模拟栈,其实就是读入出错和忘记了处理空格,改了之后,成功AC了.代码如下: #include<iostrea…

“模拟“题,运用哈希,不断地按照一定运算规律对一个结果进行计算,如果重复出现就停止并且输出该数.注意到仔细看题,这种题一定要细心! POJ - 2183 Bovine Math Geniuses Time Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u Description Farmer John loves to help the cows further their mathematical skills…

Musical Theme Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 15900   Accepted: 5494 Description A musical melody is represented as a sequence of N (1<=N<=20000)notes that are integers in the range 1..88, each representing a key on the…

题目:http://poj.org/problem?id=3080 Sample Input 3 2 GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA 3 GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA GATACTAGATACTA…

题目链接:http://poj.org/problem?id=2389 题目大意: 大数相乘. 解题思路: java BigInteger类解决 o.0 AC Code: import java.math.BigInteger; import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); while (sc.h…

题目链接:http://poj.org/problem?id=2406 题意:确定字符串最多是多少个相同的字串重复连接而成的 思路:关键是找到字符串的最小循环节 code: #include <cstdio> #include <cstring> ; char s[MAXN]; int next[MAXN]; void GetNext() { int len = strlen(s); ; ; next[] = -; while (i < len) { == j || s[i]…

题目链接: http://poj.org/problem?id=1509 题意: 求循环字符串的最小表示. 分析: 浅析"最小表示法"思想在字符串循环同构问题中的应用 判断两字符串是否是循环同构的过程就是在求字符串的最小表示,即如果两个字符串是循环同构的,那么当前两指针i=M(s1) 且j=M(s2) 的时候,一定可以得到u[i→i+s1−1] = w[j→j+s2−1] , 所以我们把给定序列看成两个循环同构的字符串,然后求一下最小表示就好了. 代码: #include<cst…

题目大意: 给定n个字符串,求出现在不小于k/2个字符串中的最长子串. 二分找对应子串长度的答案,将所有字符串链接成一个长字符串求后缀数组,记录每一个位置本属于第几个字符串,利用height查询的时候, 根据记录的位置不断判断是否出现重复的字符串是在同一个字符串内的 #include <cstdio> #include <cstring> #include <vector> #include <iostream> using namespace std; t…

题目链接:http://poj.org/problem?id=2389 题目意思:就是大整数乘法. 题目中说每个整数不超过 40 位,是错的!!!要开大点,这里我开到100. 其实大整数乘法还是第一次写 = =.......大整数加法写得比较多.百练也有一条是大整数乘法,链接如下:http://bailian.openjudge.cn/practice/2980/ 一步一步模拟即可,代码就是按这个来写的. 以 835 * 49 为例(亲爱的读者,允许我截图吧) 简直就是神奇呀----- #inc…

给出两个长度小于1000的字符串,有三种操作,插入一个字符,删除一个字符,替换一个字符. 问A变成B所需的最少操作数(即编辑距离) 考虑DP,可以用反证法证明依次从头到尾对A,B进行匹配是不会影响答案的 令dp[i][j]表示A[i]~[lenA]变成B[j]~[lenB]的最优解. 如果把B[j]插入到A[i]前,dp[i][j]=dp[i][j+1]+1 如果删除A[i],dp[i][j]=dp[i+1][j]+1. 如果A[i]==B[j], dp[i][j]=dp[i+1][j+1].…

#include<iostream>//写字符串的题目可以用这种方式:str[i][j] &str[i] using namespace std; int main() {int n,m,i,j,num,a[101],b[101],t,k; char str[101][51]; cin>>n>>m; for(i=0;i<m;i++) { cin>>str[i]; num=0; for(j=0;j<n-1;j++) for(k=j+1;k&…

Life Forms Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 16223   Accepted: 4763 Description You may have wondered why most extraterrestrial life forms resemble humans, differing by superficial traits such as height, colour, wrinkles, e…

                                                                                                  Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 38038   Accepted: 15740 Description Given two strings a and b we define a*b t…

题意:给出nn(1≤n≤50,1≤n≤50) 个病毒DNA序列,长度均不超过20.现在给出一个长度不超过1000的字符串,求至少要更换多少个字符, 才能使这个字符串不包含这些DNA序列. 析:利用前缀来做好状态转移. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #…

Period Time Limit: 3000MS Memory Limit: 30000K Description For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for e…

POJ1056 给定若干个字符串的集合 判断每个集合中是否有某个字符串是其他某个字符串的前缀 (哈夫曼编码有这个要求) 简单的过一遍Trie就可以了 #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<vector> #include<queue> #include<algor…

题意:给你一个字符串,求这个字符串到第 i 个字符为止的重复子串的个数. 解法:判断重复子串的语句很重要!!if (p && i%(i-p)==0) printf("%d %d\n",i,i/(i-p)); 我之前一直不是很理解,而实际上多枚举几种情况就好了.若是重复的,那么next[i]肯定是最大值,值余下一个循环节不同:而若不是,next[i]表示的前缀和后缀串的和重叠部分不一样以外的部分就肯定空出来,不能整除的.(P.S.我在说些什么......m(．＿．)m)…

用记录附加信息的val数组记录次数即可. trie的原理:每个可能出现的字目给一个编号c,那么整个树就是一个c叉树 ch[u][c]表示 节点u走c边过去之后的节点 PS:trie树还有种动态写法,使用指针和动态分配内存代替了连续的ch数组,更加节省内存. Reference:http://blog.csdn.net/architect19/article/details/8966247 #include <cstdio> #include <cstring> #include &…

题意:求子串在文本串中出现了多少次. 解法:使用KMP的next[ ]和tend[ ]数组计数. #include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> using namespace std; ,M=; char s[N],ss[M]; int n,m; int next[N];//,tend[M]; void kmp() { memset(next,,sizeof(n…

/* * POJ_1159.cpp * * Created on: 2013年10月29日 * Author: Administrator */ #include <iostream> #include <cstdio> #include <cmath> #include <algorithm> #include <cstring> using namespace std; const int maxn = 5005; char str1[max…

描述 Farmer John has purchased a lush new rectangular pasture composed of M by N (1 ≤ M ≤ 12; 1 ≤ N ≤ 12) square parcels. He wants to grow some yummy corn for the cows on a number of squares. Regrettably, some of the squares are infertile and can't be…

Description Given a list of phone numbers, determine if it is consistent in the sense that no number is the prefix of another. Let's say the phone catalogue listed these numbers: Emergency 911 Alice 97 625 999 Bob 91 12 54 26 In this case, it's not p…

Best Cow Line Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9284   Accepted: 2826 Description FJ is about to take his N (1 ≤ N ≤ 2,000) cows to the annual"Farmer of the Year" competition. In this contest every farmer arranges his…

Blue Jeans Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20695   Accepted: 9167 Description The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of thousa…

-----------------------------最优化问题------------------------------------- ----------------------常规动态规划  SOJ1162 I-Keyboard  SOJ1685 Chopsticks SOJ1679 Gangsters SOJ2096 Maximum Submatrix  SOJ2111 littleken bg SOJ2142 Cow Exhibition  SOJ2505 The County…