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
1
23 BRB
output
Copy
1
21 GRB
input
Copy
1
27 RGBGRBB
output
Copy
1
23 RGBRGBR
思路：一共有6种情况，选出一个改变最小的
代码：
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162#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; }
最后
以上就是傲娇月光最近收集整理的关于Codeforces-C-Nice Garland（枚举+暴力)的全部内容，更多相关Codeforces-C-Nice内容请搜索靠谱客的其他文章。
发表评论 取消回复