Nice Garland CodeForces - 1108C （思维+暴力）

57 阅读 0 评论 38 点赞

我是靠谱客的博主慈祥草莓，最近开发中收集的这篇文章主要介绍Nice Garland CodeForces - 1108C （思维+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 n

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 t

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 n

The second line of the input contains the string s

Output

In the first line of the output print one integer r

In the second line of the output print one string t

Examples

Input

3
BRB

Output

1
GRB

Input

7
RGBGRBB

Output

3
RGBRGBR

题意：简单词汇，可以自行阅读。
思路：因为题目的限制，导致如果这个字符串的长度大于等于3的时候，一定是RBG这三个字符的某一个排列的循环。
那么RBG一共最多有6种排列方式,( A(3,3) )
我们手写出这6种全排列。

{
"RGB",
"RBG",
"GRB",
"BRG",
"GBR",
"BGR"};

然后以此每一种排列去和字符串进行匹配，找出让字符串为3字符循环的最小消耗即可，
中间开一个变量k来记录是哪个排列方式让消耗最小，然后最后输出他的循环节即可。
细节看代码。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int n;
char a[maxn];
char s[50][50]=
{
"RGB",
"RBG",
"GRB",
"BRG",
"GBR",
"BGR"};
int main()
{
    gg(n);
    scanf("%s",a);
    if(n==1)
    {
        printf("0n");
        printf("%s",a);
    }else
    {
        int ans=inf;
        int cnt;
        int k;
        repd(i,0,5)
        {
            cnt=0;
            repd(j,0,n-1)
            {
                if(a[j]!=s[i][(j%3)])
                {
                    cnt++;
                }
            }
            if(cnt<ans)
            {
                ans=cnt;
                k=i;
            }
        }
        printf("%dn",ans );
        repd(i,0,n-1)
        {
            putchar(s[k][(i%3)]);
        }
        printf("n");
    }

    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == 'n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}