ZJNU 1531 - Diushou Juan - Intermediate

It may be the same number of chunks in the first array gp

For example RRGGGRBBBBRR

Then gp [1 ~ 5] = {2,3,1,4,2}

First of all know, to make the same if not adjacent, only the number of each gp [i] / 2 rounded down to obtain a minimum of change

For example RGGGR, look G, G only need to change the center

For example RGGGGR, look G, may choose to alter or G 1 3 2 4 and the position and,

 

Finally, consider the impact of end to end to form a ring of the answer

For example RRRGRRR

gp[1~3]={3,1,3}

The above statement can be obtained by the answer is 3/2 + 1/2 + 3/2 = 1 + 0 + 1 = 2

But actually connected end to end 6 sitting together R

At least three people need to change to meet the meaning of the questions

 

Also, if all of the same color

For example RRRRR

The above statement is only needed to change the 5/2 = 2

I.e., change the position of the person 2 and 4

But this will result in the 1 and 5 are connected end to end with the same color sat

Laid answer judged as required (5 + 1) in this case / 2 = 3

 1 /*
 2 Written By StelaYuri
 3 */
 4 #include<iostream>
 5 #include<algorithm>
 6 using namespace std;
 7 int main(){
 8     ios::sync_with_stdio(0);cin.tie(0);
 9     int T,t,N,m,i,gp[110],ans;
10     string s;
11     cin>>T;
12     for(t=1;t<=T;t++){
13         cin>>N>>s;
14         s=" "+s;//下标移位
15         ans=m=0;
16         for(i=1;i<=N;i++)
17             if(s[i]==s[i-1])
18                 gp[m]++;
19             else
20                 gp[++m]=1;
21         if(s[1]==s[N]&&N!=1)
22             if(m!=1){
23                 gp[1]+=gp[m];
24                 m--;
25             }
26             else
27                 gp[1]++;
28         for(i=1;i<=m;i++)
29             ans+=gp[i]/2;
30         cout<<"Case #"<<t<<":\n"<<ans<<'\n';
31     }
32     
33     return 0;
34 }