Mountain Subsequences(dp)
来源:互联网 发布:淘宝外卖粮票怎么刷 编辑:程序博客网 时间:2024/06/06 19:38
思路:
遍历两次求得每个字符左侧的递增子序列数和右侧的递减子序列数。
最后再遍历一遍每个字符左右序列数相乘。
代码:
#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int maxn = 1e5+5;const int mod = 2012;char str[maxn];int n, a[maxn], l[maxn], r[maxn], dp[maxn];int main(void){ while(cin >> n) { scanf(" %s", str); for(int i = 0; i < n; i++) a[i] = str[i]-'a'; memset(l, 0, sizeof(l)); memset(r, 0, sizeof(r)); memset(dp, 0, sizeof(dp)); for(int i = 0; i < n; i++) { for(int j = 0; j < a[i]; j++) l[i] = (l[i]+dp[j])%mod; //l[i]统计前i个能组成结尾小于a[i]的方案数 dp[a[i]] = (dp[a[i]]+l[i]+1)%mod; //dp[a[i]]统计当前位置以a[i]字符结尾的方案数 } memset(dp, 0, sizeof(dp)); for(int i = n-1; i >= 0; i--) { for(int j = 0; j < a[i]; j++) r[i] = (r[i]+dp[j])%mod; dp[a[i]] = (dp[a[i]]+r[i]+1)%mod; } int ans = 0; for(int i = 0; i < n; i++) ans = (ans+l[i]*r[i])%mod; printf("%d\n", ans); } return 0;}
Mountain Subsequences
Problem Description
Coco is a beautiful ACMer girl living in a very beautiful mountain. There are many trees and flowers on the mountain, and there are many animals and birds also. Coco like the mountain so much that she now name some letter sequences as Mountain Subsequences.
A Mountain Subsequence is defined as following:
1. If the length of the subsequence is n, there should be a max value letter, and the subsequence should like this, a1 < ...< ai < ai+1 < Amax > aj > aj+1 > ... > an
2. It should have at least 3 elements, and in the left of the max value letter there should have at least one element, the same as in the right.
3. The value of the letter is the ASCII value.
Given a letter sequence, Coco wants to know how many Mountain Subsequences exist.
Input
For each case there is a number n (1<= n <= 100000) which means the length of the letter sequence in the first line, and the next line contains the letter sequence.
Please note that the letter sequence only contain lowercase letters.
Output
Example Input
4abca
Example Output
4
Hint
aba, aca, bca, abca
- Mountain Subsequences(dp)
- sdut 2607 Mountain Subsequences (dp)
- [dp] upcpj 2221 Mountain Subsequences
- SDUT Mountain Subsequences 2607 dp(字母表优化) 2013年山东acm省赛
- sdut 2605 Mountain Subsequences(树状数组)
- 山东省第四届ACM大学生程序设计竞赛 Mountain Subsequences dp
- sdutoj-2607-Mountain Subsequences
- 第四届 Mountain Subsequences
- Mountain Subsequences(山东省第四届ACM大学生程序设计竞赛)
- Mountain Number (数位dp)
- Hrbust 2051 Mountain Subsequences【dp+思维】【哈理工OJ 800题纪念】
- [ACM] SDUT 2607 Mountain Subsequences
- 12222 - Mountain Road(dp+贪心)
- 12222 - Mountain Road(dp+贪心)
- uva 12222 - Mountain Road(dp+贪心)
- UVALive 4613 Mountain Road(DP)
- 第四届 山东省ACM SDUT 2607 Mountain Subsequences(LIS+哈希 OR 线段树 待解决)
- UVa 10069 Distinct Subsequences(大数+DP)
- DAX-PowerBI系列
- [网易云课堂]Linux内核分析(二)—— mykernel内核部署及简单时间片轮转程序分析
- leetcode:451. Sort Characters By Frequency解题报告
- “玲珑杯”ACM比赛 Round #11 D
- 看美剧如何遮住字幕
- Mountain Subsequences(dp)
- 19 简易的基于Redis的SessionManager
- Linux内核分析(二) 任务切换
- CSS水平垂直居中
- HTTPS详解
- JAVA网页图片验证码的实现
- 深入理解HTTP协议
- LINUX 下面不同主机远程拷贝文件 scp 命令的使用
- LeetCode 155. Min Stack 题解