dp[i][j]:用A的前i的字符和B的前j个字符能否组成i+j长度的合法C串 O(n^2)的方法有点糟糕 /*H E A D*/ char str1[maxn],str2[maxn],str3[maxn]; int dp[1003][1003]; int main(){ int T=read(),kase=0; while(T--){ s1(str1);s1(str2);s1(str3); memset(dp,0,sizeof dp); dp[0][0]=1; dp[1][0]=(str1[1…

目录 题目地址 题干 代码和解释 参考 题目地址 hdu1501 题干 代码和解释 最优子结构分析:设这三个字符串分别为a.b.c,如果a.b可以组成c,那么c的最后一个字母必定来自a或者b的最后一个字母.c去除最后一位,就变成由a-1和b或者a和b-1构成c-1的问题. 状态转移方程:DP[i][j]表示c中i个字符来自于a,j个字符来自于b,即由a的前i个字符和b的前j个字符组成c的前i+j个字符.DP[i][j]为1则真,为0则假. /*给3个字符串,让你判断能不能组合前两个字符串来获得第…

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…

I - 简单dp 例题扩展 Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description Nowadays, a kind of chess game called “Super Jumping! Jumping! Jumping!” is very popular in HD…

题意:给你n种花,m个盆,花盆是有顺序的,每种花只能插一个花盘i,下一种花的只能插i<j的花盘,现在给出价值,求最大价值 简单dp #include <iostream> #include<cstdio> #include<cstring> using namespace std; #define N 110 int dp[N][N],a[N][N]; int main(int argc, char** argv) { int n,m,i,j; while(sca…

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…