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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.
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.
Print the name of the winner.
3
mike 3
andrew 5
mike 2
andrew
3
andrew 3
andrew 2
mike 5
andrew
题意：给出一个各轮选手得分,最后输出分最高的那个人.
如果有多人分数一样高,那就输出先达到这个分数的人.
题解：反正就是模拟了,直接用stl的map好了.
先遍历每个人的得分 统计出最高的 在在这些人中遍历 第一个达到最高得分的人输出
第一次写map 。。可能用的不是很好。
#include <cstdio>
#include <string>
#include <map>
using namespace std;
const int maxn = 1011;
char str[maxn][36];
int score[maxn];
int main(){
int n;
scanf("%d", &n);
map<string, int> name, nametmp;
int nans = 0;
for(int i = 0; i < n; ++i){
scanf("%s%d", str[i], &score[i]);
name[str[i]] += score[i];
}
for(int i = 0; i < n; ++i){
if(name[str[i]] > nans){
nans = name[str[i]];
}
}
for(int i = 0; i < n; ++i){
if(name[str[i]] == nans){
nametmp[str[i]] += score[i];
if(nametmp[str[i]] >= nans){
puts(str[i]);
break;
}
}
}
return 0;
}
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原文地址：http://blog.csdn.net/u012431590/article/details/45836707
