Problem Description Given a matrix with n rows and m columns ( n+m ,) and you want to go to the number at bottom-right corner (n,m). And you must go right or go down every steps. Let the numbers you go through become an array a1,a2,...,a2k. The cost…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5569 简单dp,恶心的边界处理,无语: if((i+j)%2==1) dp[i][j]=a[i-1][j]*a[i][j]+min(dp[i-2][j], dp[i-1][j-1]); dp[i][j]=min(dp[i][j], a[i][j-1]*a[i][j]+min(dp[i-1][j-1], dp[i][j-2])); #include<cstdio> #include<cstri…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2571 简单dp, dp[n][m] +=(  dp[n-1][m],dp[n][m-1],d[i][k] ) k 为m的因子 PS:0边界要初始为负数(例如-123456789)越大越好 代码: #include <stdio.h> #include <string.h> int dp[25][1005]; #define max(x,y) x > y ? x : y int m…

免费馅饼 Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 102   Accepted Submission(s) : 35 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description 都说天上不会掉馅饼,但有一天gameboy正走在回家的小径上,…

建议24:边界,边界,还是边界 import java.util.Scanner; public class Client { //一个会员拥有产品的最大数量 public final static int LIMIT = 2000; public static void main(String[] args) { //会员当前拥有产品数量 int cur = 1000; Scanner input = new Scanner(System.in); System.out.print("请输入需…

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…