Codeforces 2A Winner 迭代器陷阱

最新推荐文章于 2018-03-01 17:25:13 发布

三级头不怕通

最新推荐文章于 2018-03-01 17:25:13 发布

阅读量455

点赞数

CC 4.0 BY-SA版权

分类专栏： codeforces

本文链接：https://blog.youkuaiyun.com/u011535421/article/details/79294814

codeforces 专栏收录该内容

40 篇文章

订阅专栏

该博客介绍了Berland流行的卡牌游戏Berlogging的获胜规则。当游戏结束时，如果有且仅有一个玩家拥有最大分数，则该玩家为赢家。如果多个玩家的分数相同且为最大值，则首先达到这个最大分数的玩家获胜。博客中提供了一个C++代码示例，用于根据输入的游戏回合数据找出获胜者。通过对输入的回合数据进行处理，计算每个玩家的总分数，并确定最早达到最高分的玩家作为胜者。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

A. Winner

time limit per test

1 second

memory limit per test

64 megabytes

input

standard input

output

standard output

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

#include<bits/stdc++.h> #define ll long long using namespace std; const int maxn = 10000; int main(){ int n; map<string, ll> score; cin>>n; string st[maxn]; int tmp[maxn]; for (int i=0; i<n; i++){ cin>>st[i]>>tmp[i]; score[st[i]]+=tmp[i]; } long long maxx = 0x8000000000000000; for (map<string,ll> :: iterator itr = score.begin(); itr != score.end(); itr ++){ maxx = max(itr->second, maxx); } for (map<string,ll> :: iterator itr = score.begin(); itr != score.end(); ){//这里不能再for循环里加itr++

//因为下面erase了会爆炸 if (itr->second < maxx) { score.erase(itr++); } else itr++->second = 0; } for (int i=0; i<n; i++){ if (score.find(st[i])!=score.end()){ score[st[i]] += tmp[i]; if (score[st[i]] >= maxx){ cout<<st[i]<<endl; return 0; } } } return 0; }