C. String Manipulation 1.0
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
codeforces.com/problemset/problem/91/B
Description
For example, one can change name "arca" by removing the second occurrence of character "a" to get "arc".
Polycarpus learned that some user initially registered under nickname t, where t is a concatenation of k copies of string s. Also, Polycarpus knows the sequence of this user's name changes. Help Polycarpus figure out the user's final name.
Input
The first line contains an integer k (1 ≤ k ≤ 2000). The second line contains a non-empty string s, consisting of lowercase Latin letters, at most 100 characters long. The third line contains an integer n (0 ≤ n ≤ 20000) — the number of username changes. Each of the next n lines contains the actual changes, one per line. The changes are written as "pi ci" (without the quotes), where pi (1 ≤ pi ≤ 200000) is the number of occurrences of letter ci, ci is a lowercase Latin letter. It is guaranteed that the operations are correct, that is, the letter to be deleted always exists, and after all operations not all letters are deleted from the name. The letters' occurrences are numbered starting from 1.
Output
Print a single string — the user's final name after all changes are applied to it.
Sample Input
2
bac
3
2 a
1 b
2 c
Sample Output
acb
HINT
题意
题解：
代码
//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 2000001
#define mod 1000000007
#define eps 1e-9
int Num;
char CH[];
const int inf=0x3f3f3f3f;
inline ll read()
{
int x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
} //************************************************************************************** int check[maxn];
vector<int> kiss[];
string os,s;
int main()
{
int k=read();
cin>>os;
for(int i=;i<k;i++)
s+=os;
for(int i=;i<s.size();i++)
kiss[s[i]-'a'].push_back(i);
int n=read();
for(int i=;i<n;i++)
{
int a;
char b;
cin>>a>>b;
check[kiss[b-'a'][a-]]=;
kiss[b-'a'].erase(kiss[b-'a'].begin()+a-);
}
for(int i=;i<s.size();i++)
if(!check[i])
cout<<s[i]; }