题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2227 Find the nondecreasing subsequences                                  Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)                                             …

Problem Description How many nondecreasing subsequences can you find in the sequence S = {s1, s2, s3, ...., sn} ? For example, we assume that S = {1, 2, 3}, and you can find seven nondecreasing subsequences, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1,…

http://acm.hdu.edu.cn/showproblem.php?pid=2227 用dp[i]表示以第i个数为结尾的nondecreasing串有多少个. 那么对于每个a[i] 要去找 <= a[i]的数字那些位置,加上他们的dp值即可. 可以用树状数组维护 #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <algori…

Find the nondecreasing subsequences Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1235    Accepted Submission(s): 431 Problem Description How many nondecreasing subsequences can you find in…

Find the nondecreasing subsequences Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1072    Accepted Submission(s): 370 Problem Description How many nondecreasing subsequences can you find in t…

题目大意:给定一个序列,求出其所有的上升子序列. 题解:一开始我以为是动态规划,后来发现离散后树状数组很好做,首先,c保存的是第i位上升子系列有几个,那么树状数组的sum就直接是现在的答案了,不过更新时不要忘记加1,因为当前元素本身也是一个子序列,比如数列离散后为1 3 2 4 5,那么第一位得到之前的答案为0,更新时1位加1,第二位算出为1,更新时3位加(1+1),第三位也一样,一次类推,同树状数组求逆序对的方法一样,但是更新的不是1,而是之前所有的答案数加1. #include <iostr…

Find the nondecreasing subsequences Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1844    Accepted Submission(s): 677 Problem Description How many nondecreasing subsequences can you find in t…

Description How many nondecreasing subsequences can you find in the sequence S = {s1, s2, s3, ...., sn} ? For example, we assume that S = {1, 2, 3}, and you can find seven nondecreasing subsequences, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}.…

HDU 1024 题目大意:给定m和n以及n个数,求n个数的m个连续子系列的最大值,要求子序列不想交. 解题思路:<1>动态规划,定义状态dp[i][j]表示序列前j个数的i段子序列的值,其中第i个子序列包括a[j], 则max(dp[m][k]),m<=k<=n 即为所求的结果 <2>初始状态: dp[i][0] = 0, dp[0][j] = 0; <3>状态转移: 决策:a[j]自己成为一个子段,还是接在前面一个子段的后面 方程: a[j]直接接在前面…

HDU 1231 题目大意以及解题思路见: HDU 1003题解,此题和HDU 1003只是记录的信息不同,处理完全相同. /* HDU 1231 最大连续子序列 --- 入门DP */ #include <cstdio> #include <cstring> ]; int main() { #ifdef _LOCAL freopen("D:\\input.txt", "r", stdin); #endif int n; int maxSum…