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:

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
;
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;
}