Shopping

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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Total Submission(s): 2628    Accepted Submission(s): 920



Problem Description

Every girl likes shopping,so does dandelion.Now she finds the shop is increasing the price every day because the Spring Festival is coming .She is fond of a shop which is called "memory". Now she wants to know the rank of this shop's price after the change of everyday.

 


Input

One line contians a number n ( n<=10000),stands for the number of shops.
Then n lines ,each line contains a string (the length is short than 31 and only contains lowercase letters and capital letters.)stands for the name of the shop.
Then a line contians a number m (1<=m<=50),stands for the days .
Then m parts , every parts contians n lines , each line contians a number s and a string p ,stands for this day ,the shop p 's price has increased s.

 


Output

Contains m lines ,In the ith line print a number of the shop "memory" 's rank after the ith day. We define the rank as :If there are t shops' price is higher than the "memory" , than its rank is t+1.

 


Sample Input 

3
memory
kfc
wind
2
49 memory
49 kfc
48 wind
80 kfc
85 wind
83 memory

 


Sample Output 

1
2

 


Author

dandelion

 


Source

​​曾是惊鸿照影来​​

 


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解析：直接用stl中的map暴力模拟即可。

代码：


#include<cstdio>
#include<iostream>
#include<map>
#include<cstring>
#include<string>
#define ms(a) memset(a,0,sizeof(a))
using namespace std;

const int maxn=1e4;
map<string,int> num;
int price[maxn+10],a[maxn+10];
string s,ans="memory";

int main()
{
  freopen("1.in","r",stdin);
  
  int n,m,i,j,k;
  while(cin>>n)
    {
      for(i=1;i<=n;i++)cin>>s,num[s]=i;
      cin>>m;ms(price);
      for(i=1;i<=m;i++)
        {
          for(j=1;j<=n;j++)cin>>k>>s,price[num[s]]+=k;
          for(k=0,j=1;j<=n;j++)
        if(j!=num[ans] && price[j]>price[num[ans]])k++;
          printf("%d\n",k+1);
    }
  }
  return 0;
}