# UVALive 4764 简单dp水题（也可以暴力求解）

B - Bing it

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

Description I guess most of you played cards on the trip to Harbin, but I‘m sure you have never played the following card game. This card game has N rounds and 100000 types of cards numbered from 1 to 100000. A new card will be opened when each round begins. You can ``bing" this new card. And if the card you last ``bing" is the same with this new one, you will get 1 point. You can "bing" only one card, but you can change it into a new one. For example, the order of the 4 cards is 1 3 4 3. You can ``bing" 1 in the first round and change it into 3 in the second round. You get no point in the third round, but get 1 point in the last round. Additionally, there is a special card 999. If you ``bing" it and it is opened again, you will get 3 point.

Given the order of N cards, tell me the maximum points you can get.

Input

The input file will contain multiple test cases. Each test case will consist of two lines. The first line of each test case contains one integer N(2 N 100000). The second line of each test case contains a sequence of n integers, indicating the order of cards. A single line with the number ``0" marks the end of input; do not process this case.

Output

For each input test case, print the maximum points you can get.

Sample Input

```2
1 1
5
1 999 3 3 999
0
```

Sample Output

```1
3
```
```#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;

int dp;
bool vis;
int a;

int main(){
int n;

while(scanf("%d",&n)!=EOF){
if(n==0)
break;

memset(dp,0,sizeof(dp));
memset(a,0,sizeof(a));
memset(vis,false,sizeof(vis));

for(int i=0;i<n;i++){
scanf("%d",&a[i]);
}

int sum=-1;

int point=a;
vis[point]=true;

for(int i=1;i<n;i++){
int next=a[i];
if(!vis[next]){

dp[next]=dp[point];
vis[next]=true;
}
else{
if(next==999){
dp[next]=max(dp[point],dp[next]+3);
}
else{
dp[next]=max(dp[point],dp[next]+1);
}
}

point=next;

if(dp[next]>sum)
sum=dp[next];

}
printf("%d\n",sum);
}
return 0;
}```

## ACM ：漫漫上学路 -DP -水题 CSU 1772 漫漫上学路 Time Limit: 1000MS   Memory Limit: 131072KB   64bit IO Format: %lld & %llu Submit Status Description 对于csuxushu来说,能够在CSU(California State University)上学是他一生的荣幸.CSU校园内的道路设计的十分精巧,由n+1条水平道路和n+1条竖直道路等距交错而成,充分体现了校园深厚的文化底蕴.然而不幸的是CS市每到夏季,天降大雨,