题目意思: http://acm.hdu.edu.cn/showproblem.php? pid=2601 给出一个数N,求N=i*j+i+j一共同拥有多少种方案. 题目分析: 此题直接暴力模拟就可以.仅仅是须要将上式转化为n+1=(i+1)*(j+1)进行计算就可以. AC代码: /** *hdu2601 An easy problem *意思:求n=i*j+i+j的种类 *分析:转化为n+1=(i+1)*(j+1),求解即可了 */ #include<iostream> #include&…

title: An easy problem 数学 杭电2601 tags: [数学] 题目链接 Problem Description When Teddy was a child , he was always thinking about some simple math problems ,such as "What it's 1 cup of water plus 1 pile of dough .." , "100 yuan buy 100 pig" .…

D. Easy Problem dp(有衔接关系的dp(类似于分类讨论) ) 题意 给出一个串 给出删除每一个字符的代价问使得串里面没有hard的子序列需要付出的最小代价(子序列不连续也行) 思路 要满足hard 先要满足har 要满足har 先要满足ha 一次类推 这类问题的一个共同点是要每个地方都要满足一系列前置条件才能成立也就是说有衔接关系 所以如果是构造问题 那么dp数组加一维已经满足了几个,如果是删除问题dp数组加一维 切断了哪一个即可 所以我们可以设置dp数学 dp[i][1,2,3…

Problem E Easy Problem from Rujia Liu? Though Rujia Liu usually sets hard problems for contests (for example, regional contests like Xi'an 2006, Beijing 2007 and Wuhan 2009, or UVa OJ contests like Rujia Liu's Presents 1 and 2), he occasionally sets…

An easy problem Time Limit:3000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description When Teddy was a child , he was always thinking about some simple math problems ,such as “What it’s 1 cup of water plus 1 pile of d…

Easy Problem from Rujia Liu? Though Rujia Liu usually sets hard problems for contests (for example, regional contests like Xi'an 2006, Beijing 2007 and Wuhan 2009, or UVa OJ contests like Rujia Liu's Presents 1 and 2), he occasionally sets easy probl…

An Easy Problem?! Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7837   Accepted: 1145 Description It's raining outside. Farmer Johnson's bull Ben wants some rain to water his flowers. Ben nails two wooden boards on the wall of his barn…

http://acm.hdu.edu.cn/showproblem.php?pid=5475 An easy problem Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 755 Accepted Submission(s): 431 Problem Description One day, a useless calculator was…

UVa11991 Easy Problem from Rujia Liu?  思路:  构造数组data,使满足data[v][k]为第k个v的下标.因为不是每一个整数都会出现因此用到map,又因为每个数出现次数不等可能相差很大,因此用到vector. 注意:对于数据的清空与判空不要忘记,而map在调用之前必须有map.count的检查. 代码: #include<cstdio> #include<map> #include<vector> using namespac…

An easy problem Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5475 Description One day, a useless calculator was being built by Kuros. Let's assume that number X is showed on the screen of calculator. At first,…