Recently one of my friend Tarik became a member of the food committee of an ACM regional competition. He has been given m distinguishable boxes, he has to put n types of chocolates in the boxes. The probability that one chocolate is placed in a certa…

题目链接:http://uva.onlinejudge.org/external/110/11078.pdf a[i] - a[j] 的最大值. 这个题目马毅问了我,O(n^2)超时,记忆化一下当前最大值. #include <bits/stdc++.h> using namespace std; ],n; int main() { int t; cin>>t; while(t--) { cin>>n; ;i<n;i++) { cin>>A[i]; }…

题目:题目链接 思路:预处理出l到r为回文串的子串,然后如果j到i为回文串,dp[i] = min(dp[i], dp[j] + 1) AC代码: #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> #include <vector> #include <string> #inclu…

题目传送门:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=18201 其实是一道不算难的DP,但是搞了好久,才发现原来是题目没读清楚,囧,原序列那里简直太坑了, 看了别人好多的都是用最长公共子序列,但是我用的是最长上升子序列来做,就是将原序列逐渐递增映射成递增的数列,这道题目的数据恰好符合这个条件 比如正确的序列为$$5 \ 6 \ 4  \ 1 \  3 \  2$$,我就可以将他一一映射成为 $ID[5]\righta…

/** 链接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=34398 UVA - 11584 划分字符串的回文串子串: 简单dp 题目大意: 给一个字符串, 要求把它分割成若干个子串,使得每个子串都是回文串.问最少可以分割成多少个. 定义:dp[i]表示前0~i内的字符串划分成的最小回文串个数: dp[i] = min(dp[j]+1 | j+1~i是回文串); 先预处理flag[i][j]表示以i~j内的字符串为回文串…

Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 33384    Accepted Submission(s): 15093 Problem Description Nowadays, a kind of chess game called “Super Jumping!…

题目链接:http://codeforces.com/problemset/problem/455/A 给你n个数,要是其中取一个大小为x的数,那x+1和x-1都不能取了,问你最后取完最大的和是多少. 简单dp,dp[i]表示取i时zui最大和为多少,方程为dp[i] = max(dp[i - 1] , dp[i - 2] + cont[i]*i). #include <bits/stdc++.h> using namespace std; typedef __int64 LL; ; LL a…

Problem H. ICPC QuestTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100500/attachments Description Noura Boubou is a Syrian volunteer at ACM ACPC (Arab Collegiate Programming Contest) since 2011. She graduated from Tishreen Un…

题目链接 这道题也是简单dp里面的一种经典类型,递推式就是dp[i] = min(dp[i-150], dp[i-200], dp[i-350]) 代码如下: #include<iostream> #include <stdio.h> using namespace std; ]; int main() { ; i < ; i++) dp[i] = i; ; i < ; i++) { int minn; ) dp[i] = dp[i - ]; ) dp[i] = min…

J - 简单dp Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Description It is a little known fact that cows love apples. Farmer John has two apple trees (which are conveni…