C. Nice Garland-------字符串

C. Nice Garland

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

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.

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

复制代码3 BRB

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3
BRB

output

Copy

复制代码1 GRB

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1
GRB

input

Copy

复制代码7 RGBGRBB

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7
RGBGRBB

output

Copy

复制代码3 RGBRGBRnext_permutation(a, a + 3)输出数组a的字典序的全排列。eg

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RGBRGBR


next_permutation(a, a + 3)
输出数组a的字典序的全排列。

eg

复制代码#include<iostream> #include<algorithm> using namespace std; int main() { int ans[4]={1,2,3,4}; sort(ans,ans+4); /* 这个sort可以不用，因为{1，2，3，4}已经排好序*/ do /*注意这步，如果是while循环，则需要提前输出*/ { for(int i=0;i<4;++i) cout<<ans[i]<<" "; cout<<endl; }while(next_permutation(ans,ans+4)); return 0; }

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#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
  int ans[4]={1,2,3,4};
  sort(ans,ans+4);    /* 这个sort可以不用，因为{1，2，3，4}已经排好序*/
  do                             /*注意这步，如果是while循环，则需要提前输出*/
  {
    for(int i=0;i<4;++i)
      cout<<ans[i]<<" ";
    cout<<endl;
  }while(next_permutation(ans,ans+4));
return 0;
}

 

复制代码#include<iostream> #include<algorithm> #include<string.h> #include<map> #include<stdio.h> #include<math.h> using namespace std; #define ll long long using namespace std; const int maxn = 2e5 + 10; char str[maxn], s1[maxn], s2[maxn]; int n, a[4]; int main() { cin >> n >> str; map<int, char> m; m[0] = 'R'; m[1] = 'G'; m[2] = 'B'; int ans = maxn * 2; a[1] = 1, a[2] = 2; do { s1[0] = m[a[0]]; s1[1] = m[a[1]]; s1[2] = m[a[2]]; for (int q = 3; q < n; q++) { s1[q] = s1[q - 3]; } int cnt = 0; for (int q = 0; q < n; q++) { if (s1[q] != str[q]) cnt++; } if (cnt < ans) { ans = cnt; strcpy(s2, s1); } } while (next_permutation(a, a + 3)); printf("%dn", ans); for (int i = 0; i < n; i++) printf("%c", s2[i]); return 0; }

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#include<iostream>
#include<algorithm>
#include<string.h>
#include<map>
#include<stdio.h>
#include<math.h>
using namespace std;
#define ll long long
using namespace std;
const int maxn = 2e5 + 10;
char str[maxn], s1[maxn], s2[maxn];
int n, a[4];

int main()
{
  cin >> n >> str;
  map<int, char> m;
  m[0] = 'R'; 
  m[1] = 'G'; 
  m[2] = 'B';
  int ans = maxn * 2;
  a[1] = 1, a[2] = 2;
  do 
  {
    s1[0] = m[a[0]];
    s1[1] = m[a[1]];
    s1[2] = m[a[2]];
    for (int q = 3; q < n; q++)
    {
      s1[q] = s1[q - 3];
    }
    int cnt = 0;
    for (int q = 0; q < n; q++)
    {
      if (s1[q] != str[q])
        cnt++;
    }
    if (cnt < ans)
    {
      ans = cnt;
      strcpy(s2, s1);
    }
  } while (next_permutation(a, a + 3));
  printf("%dn", ans);
  for (int i = 0; i < n; i++)
    printf("%c", s2[i]);
  return 0;
}