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)                                             …

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…

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,…

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…

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…

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}.…

题意: 给你一个序列a[],求它的不降子序列的个数 分析: dp[i]表示以i结尾不降子序列的个数,dp[i]=sum(dp[j])+1(j<i&&a[j]<=a[i]);答案就是sum(dp[i]) 但发现一个问题n很大O(n^2)肯定超时,想起前面做的两道题,用线段树优化,它是用维护的是和,可以用BIT优化,但又发现a[i]的值很大没法存,就又想到了离散化,恩这个题就解决了. #include <map> #include <set> #includ…

题目大意就是说帮你给出一个序列a,让你求出它的非递减序列有多少个. 设dp[i]表示以a[i]结尾的非递减子序列的个数,由题意我们可以写出状态转移方程: dp[i] = sum{dp[j] | 1<=j<i && a[j] <= a[i]} + 1. 这样一来这里面所有的dp[]值的和就是最后的结果. 但是这个状态转移方程很明显复杂度是O(n^2),但是n可以达到100000,很明显会超时.既然是求前导和,很明显我们就应该可以想到用树状数组(虽然我怎么也不可能想到==!)…