CodeForces 2A. Winner

C - Winner
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d
& %I64u
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2A

Appoint description:
prayerhgq (2015-07-01)System Crawler (2016-05-16)

Description

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.

Sample Input

Input

3
mike 3
andrew 5
mike 2

Output

andrew

Input

3
andrew 3
andrew 2
mike 5

Output

andrew

题意：一共有N轮比赛，接下来的每行代表这一轮比赛中，这个人得到的分数，分数可正可负，问最先到达最大分值的是谁。感觉这道题设计很不错。

刚开始用数组模拟，但是一直没想起来怎么实现找到最先达到最大分值的人，先放一放吧。看了别人的题解是用map实现的，很方便。

#include<stdio.h>
#include<string.h>
#include<map>
#include<algorithm>
#include<iostream>
using namespace std;
const int inf = 1000000000;
struct node
{
string s;
int k;
}w[1100];
map<string,int> s,m;
int main()
{
int n,i,Max;
Max=-inf;
scanf("%d",&n);
for(i=0;i<n;i++)
{
cin>>w[i].s>>w[i].k;
s[w[i].s]+=w[i].k;//求出每个人最终的得分
}
for(i=0;i<n;i++)
if(Max<s[w[i].s])
Max=s[w[i].s];//记录最终得分的最高分
for(i=0;s[w[i].s]<Max||(m[w[i].s]+=w[i].k)<Max;i++);//找到最终得分为最高分且最先不小于最高分的人即为所求
cout<<w[i].s<<endl;
//printf("%s\n",w[i].s);
return 0;
}

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