Big String

Time Limit:?1000MS		Memory Limit:?131072K
Total Submissions:?7053		Accepted:?1684

Description

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

Input

The first line of the input contains the initial string. You can assume that it is non-empty and its length does not exceed 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. If?p?is larger than the length of the string, the character is appended to the end of the string.

Q p: Query the?p-th character of the current string. The input ensures that the?p-th character exists.

All characters in the input are digits or lowercase letters of the English alphabet.

Output

For each?Q?command output one line containing only the single character queried.

Sample Input

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

Source

POJ Monthly--2006.07.30, zhucheng

題目大意?

給一個(gè)字符串，長度不超過 10⁶，有兩種操作：

??? 1、Q x(0 < x <= len(s)) 表示查詢當(dāng)前串中第x個(gè)字符

??? 2、I c x(c為字母 0 < x <= len(s)+1)表示在第x個(gè)位置插入c字符 x == len+1表示在串尾插入

操作的總數(shù)不超過 2000

做法分析

??? 塊狀鏈表裸題。詳見代碼

#include<cstdio> #include<cstring> #include<iostream> #define m(s) memset(s,0,sizeof s) using namespace std; const int N=1010; int l[N],n,m; char s[N*N],eg[N][N*3]; void insert(int x,char c){int n1=0,p1,pn=n;for(int i=1;i<=n;i++){if(n1+l[i]>=x){pn=i;break;}if(i==n) break;n1+=l[i];}p1=x-n1;l[pn]=max(p1,l[pn]+1);for(int i=l[pn];i>p1;i--) eg[pn][i]=eg[pn][i-1];eg[pn][p1]=c; } void query(int x){int n1=0,p1,pn=n;for(int i=1;i<=n;i++){if(n1+l[i]>=x){pn=i;break;}n1+=l[i];}p1=x-n1;printf("%c\n",eg[pn][p1]); } void work(){scanf("%d",&m);int len=strlen(s),ave;ave=(len+999)/1000;n=(len-1)/ave+1;for(int i=0;i<len;i++) eg[i/ave+1][i%ave+1]=s[i],l[i/ave+1]++;while(m--){char c[2];int x;scanf("%s",c);if(c[0]=='I') scanf("%s%d",c,&x),insert(x,c[0]);else scanf("%d",&x),query(x);} } int main(){while(scanf("%s",s)==1){m(l);m(eg);work();} return 0; }

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轉(zhuǎn)載于:https://www.cnblogs.com/shenben/p/6270672.html

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