Problem:
Matrix Operation

You are a student looking for a job.
Today you had an employment examination for an IT company.
They asked you to write an efficient program to perform several operations.
First, they showed you an N \times N square matrix and a list of operations.
All operations but one modify the matrix, and the last operation outputs the character in a specified cell.
Please remember that you need to output the final matrix after you finish all the operations.

Followings are the detail of the operations:

WR r c v

(Write operation) write a integer v into the cell (r,c) (1 \leq v \leq 1,000,000)

CP r1 c1 r2 c2

(Copy operation) copy a character in the cell (r1,c1) into the cell (r2,c2)

SR r1 r2

(Swap Row operation) swap the r1-th row and r2-th row

SC c1 c2

(Swap Column operation) swap the c1-th column and c2-th column

RL

(Rotate Left operation) rotate the whole matrix in counter-clockwise direction by 90 degrees

RR

(Rotate Right operation) rotate the whole matrix in clockwise direction by 90 degrees

RH

(Reflect Horizontal operation) reverse the order of the rows

RV

(Reflect Vertical operation) reverse the order of the columns

Input

First line of each testcase contains nine integers.
First two integers in the line, N and Q, indicate the size of matrix and the number of queries, respectively (1 \leq N,Q \leq 40,000).
Next three integers, A B, and C, are coefficients to calculate values in initial matrix (1 \leq A,B,C \leq 1,000,000),
and they are used as follows: A_{r,c} = (r * A + c * B) mod C where r and c are row and column indices, respectively (1\leq r,c\leq N).
Last four integers, D, E, F, and G, are coefficients to compute the final hash value mentioned in the next section (1 \leq D \leq E \leq N, 1 \leq F \leq G \leq N, E - D \leq 1,000, G - F \leq 1,000).
Each of next Q lines contains one operation in the format as described above.

Output

Output a hash value h computed from the final matrix B by using following pseudo source code.

h <- 314159265
for r = D...E
for c = F...G
h <- (31 * h + B_{r,c}) mod 1,000,000,007

where "<-" is a destructive assignment operator, "for i = S...T" indicates a loop for i from S to T (both inclusive), and "mod" is a remainder operation.

Sample Input 1

2 1 3 6 12 1 2 1 2
WR 1 1 1

Output for the Sample Input 1

676573821

Sample Input 2

2 1 3 6 12 1 2 1 2
RL

Output for the Sample Input 2

676636559

Sample Input 3

2 1 3 6 12 1 2 1 2
RH

Output for the Sample Input 3

676547189

Sample Input 4

39989 6 999983 999979 999961 1 1000 1 1000
SR 1 39989
SC 1 39989
RL
RH
RR
RV

Output for the Sample Input 4

458797120

Solution in C++11 (judge verdict: Accepted)

#include<stdio.h>
#include<algorithm>
#include<map>
using namespace std;
int row[41000];
int col[41000];
map<pair<int,int>,int> m;
char str[10];
int dir;
int n;
pair<int,int> conv(int a,int b){
	int ret1=0;
	int ret2=0;
	switch(dir&3){
		case 0:
			ret1=a;
			ret2=b;
			break;
		case 1:
			ret1=b;
			ret2=n-a+1;
			break;
		case 2:
			ret1=n-a+1;
			ret2=n-b+1;
			break;
		case 3:
			ret1=n-b+1;
			ret2=a;
			break;
	}
	if(dir&4)swap(ret1,ret2);
	return make_pair(row[ret1],col[ret2]);
}
int main(){
	int a,b,c,d,e,f,g;
	int q;
	scanf("%d%d%d%d%d%d%d%d%d",&n,&q,&a,&b,&c,&d,&e,&f,&g);
	long long h=314159265;
	long long mod=1000000007;
	for(int i=1;i<=n;i++)row[i]=i;
	for(int i=1;i<=n;i++)col[i]=i;
	while(q--){
		scanf("%s",str);
		if(str[0]=='W'){
			int R,C,V;
			scanf("%d%d%d",&R,&C,&V);
			pair<int,int> at=conv(R,C);
			m[at]=V;
		}
		if(str[0]=='C'){
			int r1,c1,r2,c2;
			scanf("%d%d%d%d",&r1,&c1,&r2,&c2);
			pair<int,int> at1=conv(r1,c1);
			pair<int,int> at2=conv(r2,c2);
			if(m.count(at1))m[at2]=m[at1];
			else m[at2]=((long long)at1.first*a+(long long)at1.second*b)%c;
		}
		if(str[0]=='S'){
			int r1,r2;
			scanf("%d%d",&r1,&r2);
			if((dir/4+(str[1]=='R')+dir%4)&1){
				if((dir+1+dir/4*1+(str[1]!='R')*2)%4<2)swap(row[r1],row[r2]);
				else swap(row[n-r1+1],row[n-r2+1]);
			}else{
				if((dir+dir/4*3+(str[1]=='R')*2)%4<2)swap(col[r1],col[r2]);
				else swap(col[n-r1+1],col[n-r2+1]);
			}
		}
		if(str[0]=='R'){
			if(str[1]=='L'){
				dir=(dir&4)+((dir%4+1)&3);
			}
			if(str[1]=='R'){
				dir=(dir&4)+((dir%4+3)&3);
			}
			if(str[1]=='H'){
				dir=((dir&4)?0:4)+(dir%4)^1;
			}
			if(str[1]=='V'){
				dir=((dir&4)?0:4)+(dir%4)^3;
			}
		}
	}/*
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			pair<int,int> at=conv(i,j);
			int ans;
			if(m.count(at))ans=m[at];
			else ans=((long long)at.first*a+(long long)at.second*b)%c;
			printf("%d ",ans);
		}
		printf("\n");
	}*/
	for(int i=d;i<=e;i++){
		for(int j=f;j<=g;j++){
			pair<int,int> at=conv(i,j);
			if(m.count(at))h=(h*31+m[at])%mod;
			else h=(h*31+((long long)at.first*a+(long long)at.second*b)%c)%mod;
		}
	}
	printf("%lld\n",h);
}