cf #2 A. Winner
2015-11-03 19:50
295 查看
http://codeforces.com/contest/2/problem/A
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.
Sample test(s)
input
output
input
output
多个球员踢球,每一次都会给其打一个分,最终得分最高者获胜,如果有多个最高分,第一个超过(!!!)最高分获胜(坑死不偿命啊);输出该球员名字
#include <iostream>
#include <map>
#include <string>
#include <string.h>
#include <set>
using namespace std;
char mm[1005][50];
char name[1005][50];
int main()
{
int n;
int const INF=100000;
int score[1005],maxn;
char str[50];
int ans[1050];
while(cin>>n)
{
maxn=-1005;
map<string,int>mp;
map<string,int>::iterator it;
set<string>Name;
for(int i=0; i<n; i++)
{
cin>>str>>score[i];
strcpy(name[i],str);
if(mp.find(str)!=mp.end())
mp[str]=mp[str]+score[i];
else
mp[str]=score[i];
}
int k=0;
for(it=mp.begin();it!=mp.end();it++)
{
if(it->second>maxn)
maxn=it->second;
}
for(int i=n-1;i>=0;i--)
{
if(mp[name[i]]==maxn&&Name.find(name[i])==Name.end())
{
strcpy(mm[k++],name[i]);
Name.insert(name[i]);
}
}
memset(ans,0,sizeof(ans));
for(int j=0;j<n;j++)
{
for(int i=0;i<k;i++)
{
if(strcmp(name[j],mm[i])==0)
{
ans[i]+=score[j];
}
if(ans[i]>=maxn)
{
cout<<mm[i]<<endl;
goto endW;
}
}
}
endW:;
}
return 0;
}
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.
Sample test(s)
input
3 mike 3 andrew 5 mike 2
output
andrew
input
3 andrew 3 andrew 2 mike 5
output
andrew
多个球员踢球,每一次都会给其打一个分,最终得分最高者获胜,如果有多个最高分,第一个超过(!!!)最高分获胜(坑死不偿命啊);输出该球员名字
#include <iostream>
#include <map>
#include <string>
#include <string.h>
#include <set>
using namespace std;
char mm[1005][50];
char name[1005][50];
int main()
{
int n;
int const INF=100000;
int score[1005],maxn;
char str[50];
int ans[1050];
while(cin>>n)
{
maxn=-1005;
map<string,int>mp;
map<string,int>::iterator it;
set<string>Name;
for(int i=0; i<n; i++)
{
cin>>str>>score[i];
strcpy(name[i],str);
if(mp.find(str)!=mp.end())
mp[str]=mp[str]+score[i];
else
mp[str]=score[i];
}
int k=0;
for(it=mp.begin();it!=mp.end();it++)
{
if(it->second>maxn)
maxn=it->second;
}
for(int i=n-1;i>=0;i--)
{
if(mp[name[i]]==maxn&&Name.find(name[i])==Name.end())
{
strcpy(mm[k++],name[i]);
Name.insert(name[i]);
}
}
memset(ans,0,sizeof(ans));
for(int j=0;j<n;j++)
{
for(int i=0;i<k;i++)
{
if(strcmp(name[j],mm[i])==0)
{
ans[i]+=score[j];
}
if(ans[i]>=maxn)
{
cout<<mm[i]<<endl;
goto endW;
}
}
}
endW:;
}
return 0;
}
相关文章推荐
- Visualizing Chess Data With ggplot
- Bit String Reordering(模拟题)
- IOS 下载、查询等
- Ubuntu 安装 vim 失败
- 使用Java来实现String字符串和Byte[]数组的转换
- socket 的用法
- windows下安装numpy模块的安装
- LeetCode——Serialize and Deserialize Binary Tree
- Android RelativeLayout布局位置属性
- javascript转到新的页面
- 测试当前机器的大小端模式(两种方法)
- 【leetcode】22. Generate Parentheses
- 如何修改Android应用包名
- artTemplate 自动化编译之tmod
- 【noip冲刺赛】:循环整数
- 64位CentOS6.5下安装hive
- Inna and New Matrix of Candies(CodeForces - 400B)
- lightoj 1224(trie)
- 醉醒间
- How GPUs Work