Description
Anatoly lives in the university dorm as many other students do. As you know, cockroaches are also living there together with students. Cockroaches might be of two colors: black and red. There are n cockroaches living in Anatoly's room.
Anatoly just made all his cockroaches to form a single line. As he is a perfectionist, he would like the colors of cockroaches in the line toalternate. He has a can of black paint and a can of red paint. In one turn he can either swap any two cockroaches, or take any single cockroach and change it's color.
Help Anatoly find out the minimum number of turns he needs to make the colors of cockroaches in the line alternate.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of cockroaches.
The second line contains a string of length n, consisting of characters 'b' and 'r' that denote black cockroach and red cockroach respectively.
Output
Print one integer — the minimum number of moves Anatoly has to perform in order to make the colors of cockroaches in the line to alternate.
Sample Input
5 rbbrr
1
5 bbbbb
2
3 rbr
0
Hint
In the first sample, Anatoly has to swap third and fourth cockroaches. He needs 1 turn to do this.
In the second sample, the optimum answer is to paint the second and the fourth cockroaches red. This requires 2 turns.
In the third sample, the colors of cockroaches in the line are alternating already, thus the answer is 0.
先涂色，计算r和b分别被涂了多少次，然后取大数为答案，因为如果一r一b被涂，就可以直接交换。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=5e2+10;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;
while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')
fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,
rx=getchar();return ra*fh;}
//#pragma comment(linker, "/STACK:102400000,102400000")
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int main()
{
char str[200010];
memset(str,0,sizeof(str));
int n,t,i,j,jiayi=0;;
int dian =0,flag;
// n = read();
// t = read();
scanf("%d %d",&n,&t);
flag = n;
//getchar();
scanf("%s",str+1);
for(i=1;i<=n;i++)
if(str[i] == '.')
dian = i;
for(i=dian+1;i<=n;i++)
{
if(str[i]>='5')
{
flag = i;
break;
}
}
if(i==n+1)
{
puts(str+1);
return 0;
}
while(t--)
{
str[flag] = '0';
for(i=flag-1;i>dian;i--)
{
str[i]++;
if(str[i]<':')
break;
if(str[i]==':')
str[i]='0';
}
if(i==dian)
{
jiayi = 1;
break;
}
if(str[i]>='5')
flag = i;
else
{
flag = i;
break;
}
}
if(jiayi)
{
for(i=dian-1;i>=0;i--)
{
str[i]++;
if(str[i]<':')
break;
if(str[i]==':')
{
str[i]='0';
}
}
if(str[0])
cout << '1';
for(i=1;i<dian;i++)
cout << str[i];
return 0;
}
for(i=1;i<=flag;i++)
{
cout << str[i];
}
return 0;
}