Shopping
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1400 Accepted Submission(s): 456
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
Recommend
yifenfei
//1218MS 1136K 754 B C++ //开始还想用排序，后来想想只用容器就够了 //代码好像也写复杂了，id不用另外记录 //在循环中每次循环比较一次就行了 #include<iostream> #include<algorithm> #include<string> #include<map> using namespace std; struct node{ int id; int s; }a[10005]; int main(void) { int n,m,p,id; string c; map<string,int>M; while(cin>>n) { M.clear(); for(int i=0;i<n;i++){ cin>>c; M[c]=i; if(c=="memory") id=i; a[i].s=0; a[i].id=i; } cin>>m; while(m--){ for(int i=0;i<n;i++){ cin>>p>>c; a[M[c]].s+=p; } int cnt=0; for(int i=0;i<n;i++) if(a[i].s>a[id].s) cnt++; printf("%d\n",cnt+1); } } return 0; }
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