/*
* Winner.cpp
*
* Created on: 2013-10-13
* Author: wangzhu
*/ /**
* 先找出所有选手的分数和中最大的分数和,之后在所有选手的分数和中看有几个是和最大的分数和相等,
* 1)、若有多个,则比较谁在比赛结束钱分数先达到最大分数和,则是赢家;
* 2)、若只有一个,则直接输出。
*/
#include<cstdio>
#include<iostream>
#include<map>
#include<string.h>
using namespace std;
#define NMAX 1010
struct Node {
int index, val;
char name[];
};
Node node[NMAX];
map<string, int> myMap;
map<string, int> myMap1;
map<string, int>::iterator mapIterator; void find(int n, int nmax) {
int m;
for (int i = ; i < n; i++) {
for (mapIterator = myMap1.begin(); mapIterator != myMap1.end();
mapIterator++) {
if ( == strcmp(node[i].name, mapIterator->first.c_str())) {
// printf("%d %d %s\n", node[i].index, node[i].val, node[i].name);
m = myMap1[mapIterator->first];
m += node[i].val;
myMap1[mapIterator->first] = m;
if (m >= nmax) {
printf("%s\n", node[i].name);
return;
}
}
}
}
} int main() {
freopen("data.in", "r", stdin);
int n, m,nmax;
string nmaxStr;
while (~scanf("%d", &n)) {
myMap.clear();
myMap1.clear(); for (int i = ; i < n; i++) {
scanf("%s%d", node[i].name, &node[i].val);
node[i].index = i;
myMap[node[i].name] +=node[i].val;
}
nmax = -;
for(mapIterator = myMap.begin();mapIterator!=myMap.end();mapIterator++){
if(nmax < mapIterator->second){
nmax = mapIterator->second;
}
}
m = ;
for(mapIterator = myMap.begin();mapIterator!=myMap.end();mapIterator++){
if(nmax == mapIterator->second){
nmaxStr = mapIterator->first;
myMap1[mapIterator->first] = ;
m ++;
}
}
//printf("%d %d\n",m,nmax);
if(m != ){
find(n,nmax);
}else{
printf("%s\n",nmaxStr.c_str());
}
}
return ;
}

codeforces Winner的更多相关文章

  1. codeforces 2A Winner (好好学习英语)

    Winner 题目链接:http://codeforces.com/contest/2/problem/A ——每天在线,欢迎留言谈论. 题目大意: 最后结果的最高分 maxscore.在最后分数都为 ...

  2. CodeForces 2A - Winner(模拟)

    题目链接:http://codeforces.com/problemset/problem/2/A A. Winner time limit per test 1 second memory limi ...

  3. CodeForces 2A Winner

    Winner Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Statu ...

  4. Codeforces Beta Round #2 A. Winner 水题

    A. Winner 题目连接: http://www.codeforces.com/contest/2/problem/A Description The winner of the card gam ...

  5. Codeforces 2A :winner

    A. Winner time limit per test 1 second memory limit per test 64 megabytes input standard input outpu ...

  6. Codeforces Beta Round #2 A. Winner

    A. Winner time limit per test 1 second memory limit per test 64 megabytes input standard input outpu ...

  7. Codeforces Gym100952 A.Who is the winner? (2015 HIAST Collegiate Programming Contest)

      A. Who is the winner?   time limit per test 1 second memory limit per test 64 megabytes input stan ...

  8. Codeforces Round #603 (Div. 2) C. Everyone is a Winner! 二分

    C. Everyone is a Winner! On the well-known testing system MathForces, a draw of n rating units is ar ...

  9. Codeforces Round #603 (Div. 2) C. Everyone is a Winner! (数学)

    链接: https://codeforces.com/contest/1263/problem/C 题意: On the well-known testing system MathForces, a ...

随机推荐

  1. Spring MVC中如何指定某个类或方法自适配地响应某个HTTP请求?

    方法已经找到,即调用AbstractHandlerMethodAdapter.handle() public final ModelAndView handle(HttpServletRequest  ...

  2. oc语言学习之基础知识点介绍(五):OC进阶

    一.点语法介绍 /* 以前封装后,要给属性赋值,必须调用方法 这样做,有两个缺点: 1.代码量多,调用方法要写的东西多. 2.看起来并不像是给属性赋值,也不像取值. 我们用点语法就可以更好的解决! 点 ...

  3. this 关键字

    导读 本文将列举C#中this关键字的用途 1.this 含义 2.用this 关键字避免参数与字段的混淆 3.用this关键字串联构造器 4.索引器 6.Visitor 模式 this 含义 C# ...

  4. asp.net 控件 导出 excel

    //导出EXCEL protected void btnDaoChu_Click(object sender, EventArgs e) { HttpContext.Current.Response. ...

  5. CAF(C++ actor framework)使用随笔(send sync_send)(二)

    a). 发完就忘, 就像上面anon_send 以及send #include <iostream> #include "caf/all.hpp" #include & ...

  6. 伪分布式环境下命令行正确运行hadoop示例wordcount

    首先确保hadoop已经正确安装.配置以及运行. 1.     首先将wordcount源代码从hadoop目录中拷贝出来. [root@cluster2 logs]# cp /usr/local/h ...

  7. hdu 5055 Bob and math problem

    先把各个数字又大到小排列,如果没有前导零并且为奇数,则直接输出.如果有前导零,则输出-1.此外,如果尾数为偶数,则从后向前找到第一个奇数,并把其后面的数一次向前移动,并把该奇数放到尾部. 值得注意的是 ...

  8. Oracle外部表详解(转载)

    (外部表创建主要注意创建目录访问权限问题.目录路径格式无空格等不相关字符,即必须是当前表访问用户可以访问:关于表中行数的限制问题,如果不加限制注意添加reject limit unlimited:表中 ...

  9. busbox编译出错,arm-linux-未找到命令

    1.问题:/opt/FriendlyARM/mini6410/linux/busybox-1.17.2/scripts/gcc-version.sh: 行 11: arm-linux-gcc: 未找到 ...

  10. QtSQL学习笔记(2)- 连接到数据库

    要使用QSqlQuery或者QSqlQueryModel访问一个数据库,首先需要创建并打开一个或多个数据库连接(database connections). 一般地,数据库连接是根据连接名(conne ...