Ignatius and the Princess IV
2016-05-05 10:46
344 查看
Ignatius and the Princess IV
Time Limit:1000MS Memory Limit:32767KB 64bit IO Format:%I64d
& %I64u
Submit Status Practice HDU
1029
Description
"OK, you are not too bad, em... But you can never pass the next test." feng5166 says.
"I will tell you an odd number N, and then N integers. There will be a special integer among them, you have to tell me which integer is the special one after I tell you all the integers." feng5166 says.
"But what is the characteristic of the special integer?" Ignatius asks.
"The integer will appear at least (N+1)/2 times. If you can't find the right integer, I will kill the Princess, and you will be my dinner, too. Hahahaha....." feng5166 says.
Can you find the special integer for Ignatius?
Input
The input contains several test cases. Each test case contains two lines. The first line consists of an odd integer N(1<=N<=999999) which indicate the number of the integers feng5166 will tell our hero. The second line contains the
N integers. The input is terminated by the end of file.
Output
For each test case, you have to output only one line which contains the special number you have found.
Sample Input
5
1 3 2 3 3
11
1 1 1 1 1 5 5 5 5 5 5
7
1 1 1 1 1 1 1
Sample Output
3
5
1
题意:多元素即在数列中出现次数多于n/2的元素
我们很容易的看出来,在一个序列中如果去掉2个不同的元素,那么原序列中的多元素,在新的序列中还是多元素,因此我们只要按照序列依次扫描,先把t赋值给result,增加个计数器,cnt = 1;然后向右扫描,如果跟result相同,则cnt++,不同,那么cnt --,这个真是我们从上面那个结论里得出的,一旦cnt == 0了,那么必定c不是多元素,这个时候把t赋值为result,cnt = 1;,重复该过程,知道结束,这个时候,result就是多元素,这个的时间复杂度为n,该题本来可以用数组保存每个元素,然后递归上述过程,可是,用数组超内存,因此我们可以直接按照上述过程计算
//用cin,cout会超时
#include<iostream>
#include<string.h>
using namespace std;
int main()
{
int n,i;
int t;
int cnt;
int result;
while(scanf("%d",&n)!=EOF)
{
cnt=0;
for(i=0;i<n;i++)
{
scanf("%d",&t);
if(cnt==0)
{
cnt=1;
result=t;
}
else
{
if(t==result)cnt++;
else cnt--;
}
}
printf("%d\n",result);
}
return 0;
}
用map水了一发。。。
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<ctime>
#define eps 1e-6
#define MAX 100005
#define INF 0x3f3f3f3f
#define LL long long
#define pii pair<int,int>
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
using namespace std;
map<int,int>mmap;
map<int,int >::iterator it;
int main()
{
int n;
int num;
while(~scanf("%d",&n))
{
mmap.clear();
for(int i=0;i<n;i++){
scanf("%d",&num);
if(mmap.end()!=mmap.find(num))
mmap[num]++;
else
mmap.insert(pii(num,1));
}
it=mmap.begin();
int res;
for(;it!=mmap.end();it++){
if(it->second>n/2)
{
res=it->first;
break;
}
}
printf("%d\n",res);
}
return 0;
}
Time Limit:1000MS Memory Limit:32767KB 64bit IO Format:%I64d
& %I64u
Submit Status Practice HDU
1029
Description
"OK, you are not too bad, em... But you can never pass the next test." feng5166 says.
"I will tell you an odd number N, and then N integers. There will be a special integer among them, you have to tell me which integer is the special one after I tell you all the integers." feng5166 says.
"But what is the characteristic of the special integer?" Ignatius asks.
"The integer will appear at least (N+1)/2 times. If you can't find the right integer, I will kill the Princess, and you will be my dinner, too. Hahahaha....." feng5166 says.
Can you find the special integer for Ignatius?
Input
The input contains several test cases. Each test case contains two lines. The first line consists of an odd integer N(1<=N<=999999) which indicate the number of the integers feng5166 will tell our hero. The second line contains the
N integers. The input is terminated by the end of file.
Output
For each test case, you have to output only one line which contains the special number you have found.
Sample Input
5
1 3 2 3 3
11
1 1 1 1 1 5 5 5 5 5 5
7
1 1 1 1 1 1 1
Sample Output
3
5
1
题意:多元素即在数列中出现次数多于n/2的元素
我们很容易的看出来,在一个序列中如果去掉2个不同的元素,那么原序列中的多元素,在新的序列中还是多元素,因此我们只要按照序列依次扫描,先把t赋值给result,增加个计数器,cnt = 1;然后向右扫描,如果跟result相同,则cnt++,不同,那么cnt --,这个真是我们从上面那个结论里得出的,一旦cnt == 0了,那么必定c不是多元素,这个时候把t赋值为result,cnt = 1;,重复该过程,知道结束,这个时候,result就是多元素,这个的时间复杂度为n,该题本来可以用数组保存每个元素,然后递归上述过程,可是,用数组超内存,因此我们可以直接按照上述过程计算
//用cin,cout会超时
#include<iostream>
#include<string.h>
using namespace std;
int main()
{
int n,i;
int t;
int cnt;
int result;
while(scanf("%d",&n)!=EOF)
{
cnt=0;
for(i=0;i<n;i++)
{
scanf("%d",&t);
if(cnt==0)
{
cnt=1;
result=t;
}
else
{
if(t==result)cnt++;
else cnt--;
}
}
printf("%d\n",result);
}
return 0;
}
#include<iostream> #include<string.h> using namespace std; int main() { int n,num; int cnt, result; while(~scanf("%d",&n)) { cnt=0; for(int i=0;i<n;i++) { scanf("%d",&num); if(cnt==0) cnt=1,result=num; else { if(num==result)cnt++; else cnt--; } } printf("%d\n",result); } return 0; }
用map水了一发。。。
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<ctime>
#define eps 1e-6
#define MAX 100005
#define INF 0x3f3f3f3f
#define LL long long
#define pii pair<int,int>
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
using namespace std;
map<int,int>mmap;
map<int,int >::iterator it;
int main()
{
int n;
int num;
while(~scanf("%d",&n))
{
mmap.clear();
for(int i=0;i<n;i++){
scanf("%d",&num);
if(mmap.end()!=mmap.find(num))
mmap[num]++;
else
mmap.insert(pii(num,1));
}
it=mmap.begin();
int res;
for(;it!=mmap.end();it++){
if(it->second>n/2)
{
res=it->first;
break;
}
}
printf("%d\n",res);
}
return 0;
}
相关文章推荐
- 删除文档中低于某个时间段的文档
- 数据库原理
- 一个靠谱的国外maven镜像地址
- 刚刚发现了一个快捷键 Shift+Insert
- C++第五次上级实验
- CentOS7 安装 PostGIS方法(适合国内网络
- iOS-事件的传递与响应
- 强大的PHP生成缩略图函数
- android studio生成jar包
- 深入理解Linux内核day08--进程线性地址空间
- Android Q&A | No orientation specified, and the default is horizontal,
- MYSQL使用正则表达式过滤数据
- Netflix开源技术介绍
- sox 语音格式及播放软件goldwav
- HDU 1405 The Last Practice(暴力枚举)
- Java中int转String 和 String转int 各方法效率对比
- Realm 数据库取代sqlite?
- 设计模式之装饰器模式
- Linux基本监控项目
- ios - NSTimer中target的self是强引用问题