poj 2887 块状数组

2024-04-20 05:58

文章标签 数组 poj 块状 2887

本文主要是介绍poj 2887 块状数组，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

Big String

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 7116		Accepted: 1701

Description

You are given a string and supposed to do some string manipulations(操纵).

Input

The first line of the input(投入) contains the initial string. You canassume(承担) that it is non-empty and its length does notexceed(超过) 1,000,000.

The second line contains the number of manipulation commands N (0 < N ≤ 2,000). The following N lines describe a command each. The commands are in one of the two formats below:

I ch p: Insert a character ch before the p-th character of the current string. Ifp is larger than the length of the string, the character is appended(附加) to the end of the string.
Q p: Query the p-th character of the current string. The input(投入) ensures(保证) that thep-th character exists.

All characters in the input(投入) aredigits(数字) orlowercase(小写字母) letters of the Englishalphabet(字母表).

Output

For each Q command output(输出) one line containing only the single character queried.

Sample Input

题意：

有一串字符串需要进行一些操作；当输入Q i时将第i个字符输出

当输入 I ch i 时将第 i个字符前添加一个字符ch；

这道题本来想用线段树来做的后来感觉块状数组更简单一些。

ab
7
Q 1
I c 2
I d 4
I e 2
Q 5
I f 1
Q 3

Sample Output

a
d
e

#include<string.h>
#include<stdio.h>
#include<algorithm>
#define maxn 1001
#define maxl 2000005
#define N 1001
using namespace std;
char c[maxn*maxn];
char a[maxn][maxn*3];
int len[N];
int alen;
char query(int k){int cou=0;for(int i=0;i<alen;i++){if(k<=cou+len[i]){return a[i][k-cou-1];}cou+=len[i];}
}
void add(int k,char ch){int cou=0;int r;for(int i=0;i<alen;i++){if(k<=cou+len[i]){r=i;break;}if(i==alen-1){r=alen-1;break;}cou+=len[i]; }int pos=k-cou-1;if(pos>=len[r])pos=len[r];for(int i=len[r];i>pos;i--){a[r][i]=a[r][i-1];}a[r][pos]=ch;len[r]++;}int main(){while(~scanf("%s",c)){int  slen=strlen(c);int along;memset(len,0,sizeof(len));along=(slen+999)/1000;alen=(slen+along-1)/along;for(int i=0;i<alen;i++)len[i]=along;for(int i=0;i<slen;i++){a[i/along][i%along]=c[i];}len[alen-1]=(slen-1)%along+1;int n;scanf("%d",&n);while(n--){char op[3];int k;scanf("%s",op);if(op[0]=='Q'){scanf("%d",&k);printf("%c\n",query(k));} else{scanf("%s %d", op, &k); add(k,op[0]);}}
}
return 0;
}

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