題干：

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

題目大意：

給出一些名字和分?jǐn)?shù)，正的表示加分，負(fù)的表示減分，要求你求出最大分?jǐn)?shù)而且最先得到這個(gè)最高分的那個(gè)人的名字

解題報(bào)告：

? 這題坑還是很多的。。你不能在線維護(hù)一個(gè)最大值，因?yàn)檫@個(gè)最大值還可能減下來(lái)，，你需要看的是最后比賽完了以后的最大值，記錄那個(gè)最大值。并且最后判斷輸出的時(shí)候也不能直接找到第一個(gè)大于maxx的就停止搜索了，因?yàn)檫@個(gè)人也可能后來(lái)有負(fù)分導(dǎo)致最終分?jǐn)?shù)小于maxx。

AC代碼：

#include<bits/stdc++.h>using namespace std; const int INF = 0x3f3f3f3f; map<string, int> mp; int main() {int n,sc[1000 + 5];//sc代表分?jǐn)?shù) char s[1000 + 5][50];int maxx = -INF;cin>>n;for(int i = 1; i<=n; i++) {scanf("%s %d",s[i],&sc[i]);mp[s[i]]+=sc[i]; // maxx =max(maxx,mp[s[i]]); }for(int i = 1; i<=n; i++) {maxx = max(maxx,mp[s[i]]);}map<string,int > mpp;for(int i = 1; i<=n; i++) {mpp[s[i]] +=sc[i];if(mpp[s[i]] >= maxx &&mp[s[i]] == maxx ) {printf("%s\n",s[i]);break; } }return 0 ;}

?

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