sdut 2605 Mountain Subsequences（树状数组）

Mountain Subsequences

Time Limit: 1000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描述

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.

输入

Input contains multiple test cases.

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. 

输出

For each case please output the number of the mountain subsequences module 2012.

示例输入

4abca

示例输出

4

提示

The 4 mountain subsequences are:

aba, aca, bca, abca

题目链接：http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2607

思路：看作是是任意字符，则想成了是树状数组。

刚刚好，复习一下树状数组，O(∩_∩)O~~

这个是由树状数组优化：

原先的状态转移方程为：

for(int i=0;i<n;i++){    for(int j=0;j<i;j++)    if(num[j]<num[j])       dp[i]+=dp[j];    l[i]=dp[i];    dp[i]++;}

明显的是这个时间复杂度是n^2,绝逼是要超时，gg

此时就需要用树状数组优化内层循环；

这里就不多说了，如果不知道，那么自己去了解吧b(￣▽￣)d

代码：

#include<stdio.h>#include<string.h>#include<iostream>using namespace std;const int maxn=111111;const int mod=2012;typedef long long ll;int l[maxn],r[maxn],bit[maxn],num[maxn];char s[maxn];int lowbit(int x){    return x&(-x);}void add(int x,int k){    while(x<1000)    {        bit[x]+=k;        bit[x]%=mod;        x+=lowbit(x);    }}int sum(int x){    int ans=0;    while(x)    {        ans+=bit[x];        ans%=mod;        x-=lowbit(x);    }    return ans%mod;}int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        scanf("%s",s);        memset(bit,0,sizeof(bit));        for(int i=0;i<n;i++)        {            num[i]=s[i]-' '+10;            l[i]=sum(num[i]);            add(num[i]+1,l[i]+1);        }        memset(bit,0,sizeof(bit));        for(int i=n-1;i>=0;i--)        {             r[i]=sum(num[i]);            add(num[i]+1,r[i]+1);        }        int ans=0;        for(int i=0;i<n;i++)        {            printf("%d %d\n",l[i],r[i]);            l[i]%=mod;            r[i]%=mod;            ans+=l[i]*r[i]%mod;            ans%=mod;        }        cout<<(ans+mod)%mod<<endl;    }    return 0;}