題目

Nothing is eternal in the world, Kostya understood it on the 7-th of January when he saw partially dead four-color garland.

Now he has a goal to replace dead light bulbs, however he doesn't know how many light bulbs for each color are required. It is guaranteed that for each of four colors at least one light is working.

It is known that the garland contains light bulbs of four colors: red, blue, yellow and green. The garland is made as follows: if you take any four consecutive light bulbs then there will not be light bulbs with the same color among them. For example, the garland can look like "RYBGRYBGRY", "YBGRYBGRYBG", "BGRYB", but can not look like "BGRYG", "YBGRYBYGR" or "BGYBGY". Letters denote colors: 'R' — red, 'B' — blue, 'Y' — yellow, 'G' — green.

Using the information that for each color at least one light bulb still works count the number of dead light bulbs of each four colors.

Input

The first and the only line contains the string s (4?≤?|s|?≤?100), which describes the garland, the i-th symbol of which describes the color of the i-th light bulb in the order from the beginning of garland:

'R' — the light bulb is red,
'B' — the light bulb is blue,
'Y' — the light bulb is yellow,
'G' — the light bulb is green,
'!' — the light bulb is dead.

The string s can not contain other symbols except those five which were described.

It is guaranteed that in the given string at least once there is each of four letters 'R', 'B', 'Y' and 'G'.

It is guaranteed that the string s is correct garland with some blown light bulbs, it means that for example the line "GRBY!!!B" can not be in the input data.

Output

In the only line print four integers kr,?kb,?ky,?kg — the number of dead light bulbs of red, blue, yellow and green colors accordingly.

分析

我的理解就是4中顏色的燈妹沙，找一個(gè)符合他給的序列的循環(huán)格式墩剖，因?yàn)楸ＷC每個(gè)燈都出現(xiàn)一次界轩，所以直接遍歷一邊就可以找出4個(gè)燈的循環(huán)格式反症，然后遍歷第二遍去找！出現(xiàn)的地方應(yīng)該亮什么燈存捺，統(tǒng)計(jì)一下槐沼。

ac代碼

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>

using namespace std;
//const int maxs = 1e2 + 5;
char c[4] = {'R', 'B', 'Y', 'G'};
int num_col[4];
int ind[4];
string str;
int main(){
    cin >> str;
    int size = str.size();
    for(int i = 0; i < 4; ++i){
        for(int j = i; j < size; j += 4){
            if(str[j] == '!') {
                continue;
            }
            for(int k = 0; k < 4; ++k){
                if(str[j] == c[k]){
                    ind[i] = k;
                }
            }
        }
    }
    for(int i = 0; i < 4; ++i){
        for(int j = i; j < size; j += 4){
            if(str[j] == '!'){
                ++num_col[ind[i]];
            }   
        }
    }
    bool t_peace = 1;
    for(int i = 0; i < 4; ++i){
        if(t_peace){
            t_peace = 0;
        }
        else cout << " ";
        cout << num_col[i];
    }
    return 0;
}