1 second
256 megabytes
standard input
standard output
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.
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.
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.
3
BRB
1
GRB
7
RGBGRBB
3
RGBRGBR 这个是一个思维题，需要我们找找规律，我自己没有找到。，看了题解才会的。
这个对3取模和这里有三个数字有着莫大的联系，它的这个要求，决定了这个排列应该是对这三个数的全排的一个循环，所以这个时候就只要一一比对，找到消耗最小的
就可以了。
#include<iostream>
#include<cstdio>
#include<cstring>
#include <vector>
#include <algorithm>
#include <string>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 100;
char s[maxn];
int main()
{
int n,mark=0;
cin >> n;
cin >> s;
int ans = inf;
int len = strlen(s);
string a[6] = { "RBG","RGB","BRG","BGR","GBR","GRB" };
for(int i=0;i<6;i++)
{
int cnt = 0;
for(int j=0;j<len;j++)
{
if (a[i][j % 3] != s[j]) cnt++;
}
if(cnt<ans)
{
mark = i;
ans = cnt;
}
}
printf("%d\n", ans);
for(int i=0;i<len;i++)
{
printf("%c", a[mark][i % 3]);
}
printf("\n");
return 0;
}