CF 2A Winner（STL+模拟）

Posted 2020-10-18

题目链接：http://codeforces.com/contest/2/problem/A

题目：

The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player‘s name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to m) at the end of the game, than wins the one of them who scored at least m points first. Initially each player has 0 points. It‘s guaranteed that at the end of the game at least one player has a positive number of points.

Input

The first line contains an integer number n (1  ≤  n  ≤  1000), n is the number of rounds played. Then follow n lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.

Output

Print the name of the winner.

Examples

input

3
mike 3
andrew 5
mike 2

output

andrew

input

3
andrew 3
andrew 2
mike 5

output

andrew

 题意：一堆人玩游戏，每次（总共n次）给出某个人的得分。n次游戏结束后，分数最高的那个人获胜，如果最高分数不止一人，那么最先获得大于等于最高分数者获胜，模拟一下过程就好了。挺有意思的一道小题目，这里用了map，感觉整个简洁多了。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 typedef long long LL;
 5 const int INF=0x3f3f3f3f;
 6 map <string,int> m;
 7 map <string,int> ans;
 8 string s[1234];
 9 int x[1234];
10 
11 int main(){
12     ios::sync_with_stdio(false);
13     int n,MAX=-1;
14     cin>>n;
15     for(int i=1;i<=n;i++){
16         cin>>s[i]>>x[i];
17         m[s[i]]+=x[i];
18     }
19     for(int i=1;i<=n;i++) MAX=max(MAX,m[s[i]]);
20     for(int i=1;i<=n;i++){
21         ans[s[i]]+=x[i];
22         if(m[s[i]]==MAX&&ans[s[i]]>=MAX) {cout<<s[i]<<endl;return 0;}
23     }
24     return 0;
25 }