原题链接：[http://47.121.118.174/p/618]
没人写题解我来水一篇()
注意到 $n,m$ 的取值范围，直接暴力枚举的时间复杂度是$O(n^3m^3)$会超时，所以考虑用前缀和做。
创建三个二维前缀和数组
$sumR,sumRO,sumROG，$
分别表示在$(1,1)$到$(i,j)$范围内
$R,RO,ROG$的数量。
大致的思路：遍历字符矩阵，前缀和数组先继承之前的数据
（二维前缀和模板:

$$sumR[i][j] = sumR[i-1][j]+sumR[i][j-1]-sumR[i-1][j-1];$$
），然后根据$a[i][j]$的值，对前缀和数组进行处理:
$$a[i][j]=='R'->sumR[i][j]++;$$ $$a[i][j]=='O'->sumRO[i][j]+=sumR[i][j]$$ $$a[i][j]=='G'->sumROG[i][j]+=sumRO[i][j]$$

Java代码如下：

import java.io.*;
import java.util.*;
public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
        int n = Integer.parseInt(st.nextToken()); 
        int m = Integer.parseInt(st.nextToken());
        char[][] a = new char[n+1][m+1];
        long[][] sumR = new long[n+1][m+1];
        long[][] sumRO = new long[n+1][m+1];
        long[][] sumROG = new long[n+1][m+1];
        for(int i = 1;i <= n;i++) {
        	StringBuilder sb = new StringBuilder(" "+br.readLine());
        	a[i] = sb.toString().toCharArray();
        }
        for(int i = 1;i <= n;i++) {
        	for(int j = 1;j <= m;j++) {
        		sumR[i][j] = sumR[i-1][j]+sumR[i][j-1]-sumR[i-1][j-1]+(a[i][j]=='R'?1:0);
        		sumRO[i][j] = sumRO[i-1][j]+sumRO[i][j-1]-sumRO[i-1][j-1]+(a[i][j]=='O'?sumR[i][j]:0);
        		sumROG[i][j] = sumROG[i-1][j]+sumROG[i][j-1]-sumROG[i-1][j-1]+(a[i][j]=='G'?sumRO[i][j]:0);
        	}
        }
        long ans = sumROG[n][m] % 998244353;
        pw.println(ans);
        pw.flush();
    }
}