poj 2887 块状数组

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:

1. 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.
2. 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;
}