Codeforces 2A Winner (map运用)

Winner

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on CodeForces. Original ID: 2A
64-bit integer IO format: %I64d      Java class name: (Any)

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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

Source

Codeforces Beta Round #2

题意:大概就是说一个报数的游戏,没人报一个数字,游戏结束后数字之和最大的获胜,如果相同,则先报的胜利.

此题可用数组模拟,但是有点麻烦,用STL里的map可以很快解决此问题.

AC代码:

#include <iostream>#include <cstring>#include <map>using namespace std;map<string,int>m1,m2;string name[1010];int score[1010];int main(){    int n;    cin>>n;    for(int i=0; i<n; i++)    {        cin>>name[i]>>score[i];        m1[name[i]]+=score[i];    }    int maxx=-1005;    for(int i=0;i<n;i++)    {        if(m1[name[i]]>maxx)            maxx=m1[name[i]];    }    for(int i=0;i<n;i++)    {        m2[name[i]]+=score[i];        if(m2[name[i]]>=maxx&&m1[name[i]]>=maxx)        {            cout<<name[i]<<endl;            break;        }    }    return 0;}

网上看到了q神的代码,菜鸟要学习一下

#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<iostream>#include<algorithm>#include<map>using namespace std;const int MAXN=1005;string s[MAXN];int c[MAXN];map<string,int>mp;int main(){    ios::sync_with_stdio(false);    int n;    cin>>n;    for(int i=0; i<n; i++)        cin>>s[i]>>c[i];    for(int i=0; i<n; i++)        mp[s[i]]+=c[i];    int res=0;    for(map<string,int>::iterator itr=mp.begin(); itr!=mp.end(); itr++)        res=max(res,itr->second);    for(map<string,int>::iterator itr=mp.begin(); itr!=mp.end();)    {        if(itr->second !=res)            mp.erase(itr++);//函数删除在pos位置的元素        else            (itr++)->second =0;    }    for(int i=0; i<n; i++)        if(mp.find(s[i])!=mp.end())        {            mp[s[i]]+=c[i];            if(mp[s[i]]>=res)            {                cout<<s[i]<<endl;                return 0;            }        }    return 0;}