Codeforces 758B-Blown Garland

最新推荐文章于 2019-05-15 23:33:20 发布

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最新推荐文章于 2019-05-15 23:33:20 发布

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本文介绍了一种修复部分损坏的四色灯串的方法。通过分析灯串的颜色排列规律及已知条件，确保修复后的灯串中任意连续四个灯都不相同。文章详细解释了如何确定各颜色灯的数量差异。

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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.

Examples

input
RYBGRYBGR

output
0 0 0 0

input
!RGYB

output
0 1 0 0

input
!!!!YGRB

output
1 1 1 1

input
!GB!RG!Y!

output
2 1 1 0

Note

In the first example there are no dead light bulbs.

In the second example it is obvious that one blue bulb is blown, because it could not be light bulbs of other colors on its place according to the statements.

题意：一共有四种颜色的灯RBYG，现在有！处表示这个位子的灯泡坏掉了，我们现在需要在！处放置四种颜色的灯泡，使得最终的序列，保证每连续的四个灯泡都具有四种不同的颜色。问我们需要添加多少个R,B,Y,G。保证答案唯一。

解题思路：既然保证答案唯一，那么说明每种颜色的灯的相对位置是固定的，找到每种颜色的灯出现了几次，应该出现几次，它们的差就是答案

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

using namespace std;

#define LL long long

int main()
{
    char ch[200];
    int res[4],tot[4],cnt[4];
    while(~scanf("%s",ch))
    {
        int len=strlen(ch);
        memset(tot,0,sizeof tot);
        memset(cnt,0,sizeof cnt);
        for(int i=0; i<len; i++)
        {
            tot[i%4]++;
            if(ch[i]=='R') res[0]=i%4,cnt[0]++;
            if(ch[i]=='B') res[1]=i%4,cnt[1]++;
            if(ch[i]=='Y') res[2]=i%4,cnt[2]++;
            if(ch[i]=='G') res[3]=i%4,cnt[3]++;
        }
        for(int i=0; i<3; i++)
            printf("%d ",tot[res[i]]-cnt[i]);
        printf("%d\n",tot[res[3]]-cnt[3]);
    }
    return 0;
}