Codeforces-C-Nice Garland（枚举+暴力)

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我是靠谱客的博主傲娇月光，最近开发中收集的这篇文章主要介绍Codeforces-C-Nice Garland（枚举+暴力)，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

You have a garland consisting of nn lamps. Each lamp is colored red, green or blue. The color of the ii-th lamp is sisi ('R', 'G' and 'B' — colors of lamps in the garland).

You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is nice.

A garland is called nice if any two lamps of the same color have distance divisible by three between them. I.e. if the obtained garland is tt, then for each i,ji,j such that ti=tjti=tj should be satisfied |i−j| mod 3=0|i−j| mod 3=0. The value |x||x| means absolute value of xx, the operation x mod yx mod y means remainder of xx when divided by yy.

For example, the following garlands are nice: "RGBRGBRG", "GB", "R", "GRBGRBG", "BRGBRGB". The following garlands are not nice: "RR", "RGBG".

Among all ways to recolor the initial garland to make it nice you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them.

Input

The first line of the input contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of lamps.

The second line of the input contains the string ss consisting of nn characters 'R', 'G' and 'B' — colors of lamps in the garland.

Output

In the first line of the output print one integer rr — the minimum number of recolors needed to obtain a nice garland from the given one.

In the second line of the output print one string tt of length nn — a nice garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them.

Examples

input

Copy

3
BRB

output

Copy

1
GRB

input

Copy

7
RGBGRBB

output

Copy

3
RGBRGBR

思路：一共有6种情况，选出一个改变最小的

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

char a[200005];
char b[200005];
char c[200005];
int main() {
	int n;
	cin>>n;
	scanf("%s",b);
	int len=strlen(b);
	strcpy(a,b);

	int maxn=999999;

	int s=0;


	for(int t=0; t<len; t+=3) {
		if(a[t]!='R'&&t<len) {
			a[t]='R';
			s++;
		}
		if(a[t+1]!='G'&&t+1<len) {
			a[t+1]='G';
			s++;
		}
		if(a[t+2]!='B'&&t+2<len) {
			a[t+2]='B';
			s++;
		}
	}

	if(s<maxn) {
		maxn=s;
		strcpy(c,a);

	}
	strcpy(a,b);
	s=0;
	for(int t=0; t<len; t+=3) {
		if(a[t]!='R'&&t<len) {
			a[t]='R';
			s++;
		}
		if(a[t+1]!='B'&&t+1<len) {
			a[t+1]='B';
			s++;
		}
		if(a[t+2]!='G'&&t+2<len) {
			a[t+2]='G';
			s++;

		}

	}

	if(s<maxn) {
		maxn=s;
		strcpy(c,a);

	}
	strcpy(a,b);
	s=0;

	for(int t=0; t<len; t+=3) {
		if(a[t]!='G'&&t<len) {
			a[t]='G';
			s++;
		}
		if(a[t+1]!='R'&&t+1<len) {
			a[t+1]='R';
			s++;
		}
		if(a[t+2]!='B'&&t+2<len) {
			a[t+2]='B';
			s++;
		}
	}

	if(s<maxn) {
		maxn=s;
		strcpy(c,a);

	}
	strcpy(a,b);
	s=0;

	for(int t=0; t<len; t+=3) {
		if(a[t]!='G'&&t<len) {
			a[t]='G';
			s++;
		}
		if(a[t+1]!='B'&&t+1<len) {
			a[t+1]='B';
			s++;
		}
		if(a[t+2]!='R'&&t+2<len) {
			a[t+2]='R';
			s++;
		}
	}


	if(s<maxn) {
		maxn=s;
		strcpy(c,a);

	}

	strcpy(a,b);
	s=0;
	for(int t=0; t<len; t+=3) {
		if(a[t]!='B'&&t<len) {
			a[t]='B';
			s++;
		}
		if(a[t+1]!='G'&&t+1<len) {
			a[t+1]='G';
			s++;
		}
		if(a[t+2]!='R'&&t+2<len) {
			a[t+2]='R';
			s++;
		}
	}

	if(s<maxn) {
		maxn=s;
		strcpy(c,a);

	}
	strcpy(a,b);
	s=0;
	for(int t=0; t<len; t+=3) {
		if(a[t]!='B'&&t<len) {
			a[t]='B';
			s++;
		}
		if(a[t+1]!='R'&&t+1<len) {
			a[t+1]='R';
			s++;
		}
		if(a[t+2]!='G'&&t+2<len) {
			a[t+2]='G';
			s++;
		}
	}
	if(s<maxn) {
		maxn=s;
		strcpy(c,a);

	}

	cout<<maxn<<endl;
	puts(c);
	return 0;
}