POJ 3735 Training Little cats__ matrix fast Power

The main effect of the topic:

Fesse's pet cat just gave birth to a litter of kittens, considering their health, Fesse decided to let them do more exercise. It has designed a series of actions to let them do, and let you supervise them, just in case they are too much hung up, a total of three actions, formalized description:

G I: Let the first cat get a peanut

E i: Let the first cat eat all the peanuts in its hand

S I j: let the first cat and the J Cat Exchange The Peanuts in hand

Fesse will give some of the above action (number fixed), and then repeat some of the column actions several times.

There are a number of examples first, in each case there will be N cats (1≤n≤100, numbered 1-n, measured with n as 0), repeated m (0≤m≤1,000,000,000), and the number of repetitions of each series of actions K (1≤k≤100), which is now calculated in each case, repeating all moving How many peanuts are there in the hands of each cat?

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Comment Code:

* * Problem id:poj 3735 Training Little Cats * Author:lirx.t.una                          
* LANGUAGE:C * Run time:110 MS * Run memory:416 KB
* * #include <string.h> #include <stdio.h>//maximum Number of cats//cats, due to the width of the transfer matrix 1 larger than the number of cats

The subscript is therefore marked with 1 #define MAXCATN 102 typedef long LLG;
The body of main matrix, the main matrices//The results of the original series operations Llg B[maxcatn][maxcatn];
Answer matrix, storing matrices of fast power results llg A[maxcatn][maxcatn];

Temporary matrix, storing intermediate result matrices llg T[maxcatn][maxcatn]; int Ms;//matrix size, width of the matrix (cat number plus 1) void Mul (LLG (*M1) [Maxcatn], LLG (*M2) [Maxcatn]) {//multiply//matrix multiplication, the result is stored in the first matrix, so You can avoid a lot of duplication when the return value//is passed and the time to allocate space when creating matrices//matrix1,matrix2 int i, J, k;//count variable memset (t, 0, sizeof (t));//use T to store multiplication results fo R (i = 1; I <= ms; i++) for (k = 1; k <= ms; k++) if (m1[i][k))//important pruning, otherwise tle for (j = 1; J <= ms ; J + +) T[i][j] + = m1[i][K] * M2[k][j]; MEMCPY (M1, T, sizeof (T))//overwrite result to first matrix} void Fstxp (LLG (*m) [Maxcatn], Llg N) {//fast exponentiation//Matrix fast power, algorithm and exponentiation modulo operation

	In the same way, the binary int i is adopted;

	memset (A, 0, sizeof (a));

		for (i = 1; I <= ms; i++) A[i][i] = 1ll;//First initializes answer to the unit matrix while (n) {if (n & 1) Mul (A, m);

		Mul (M, m);
	n >>= 1;
	int main () {int n, m, k;//cat number, repeat number, each cycle operand char cmd;
	int x, y;
	int i;

		LLG Swp;//swap, as Exchange meta while (scanf ("%d%d%d", &n, &m, &k), n) {ms = n + 1;

		memset (b, 0, sizeof (b)); for (i = 1; I <= ms; i++) B[i][i] = 1ll;//First initializes the primary matrix to the unit matrix//Last row as the cat Peanut record row while (k--) {scanf ("\n%c"),
			&cmd);
					Switch (cmd) {case ' G '://+ 1 only acts on the last line scanf ("%d", &x);
					b[ms][x]++;

				Break
					Case ' E '://Clear 0 acts on each line scanf ("%d", &x);
					for (i = 1; I <= ms; i++) b[i][x] = 0LL;

				Break
					Default://Exchange functions on each line scanf ("%d%d", &x, &y); for (i = 1; i; = ms;
						i++) {SWP = b[i][x];
						B[I][X] = B[i][y];
					B[i][y] = SWP;
			} break;

		} fstxp (b, M);
		for (i = 1; I <= n; i++) printf ("%lld", A[ms][i));
	Putchar (' \ n ');
return 0; }

No comment code:

#include <string.h> #include <stdio.h> #define MAXCATN 102 typedef long LLG;
LLG B[maxcatn][maxcatn];
LLG A[maxcatn][maxcatn];

LLG T[maxcatn][maxcatn];

int MS;

	void Mul (LLG (*M1) [Maxcatn], LLG (*M2) [Maxcatn]) {int I, j, K;

	memset (t, 0, sizeof (t));
					for (i = 1; I <= ms; i++) for (k = 1; k <= ms; k++) if (m1[i][k)) for (j = 1; J <= ms; j)

	T[I][J] + = m1[i][k] * M2[k][j];
MEMCPY (M1, T, sizeof (t));

	} void Fstxp (LLG (*m) [Maxcatn], Llg n) {int i;

	memset (A, 0, sizeof (a));

	for (i = 1; I <= ms; i++) A[i][i] = 1LL;

		while (n) {if (n & 1) Mul (A, m);

		Mul (M, m);
	n >>= 1;
	int main () {int n, m, K;
	char cmd;
	int x, y;
	int i;

	LLG SWP;

		while (scanf ("%d%d%d", &n, &m, &k), n) {ms = n + 1;

		memset (b, 0, sizeof (b));

		for (i = 1; I <= ms; i++) B[i][i] = 1LL;
			while (k--) {scanf ("\n%c", &cmd); Switch (cmd) {case 'G ': scanf ("%d", &x);
					b[ms][x]++;

				Break
					Case ' E ': scanf ("%d", &x);
					for (i = 1; I <= ms; i++) b[i][x] = 0LL;

				Break
					DEFAULT:SCANF ("%d%d", &x, &y);
						for (i = 1; I <= ms; i++) {SWP = b[i][x];
						B[I][X] = B[i][y];
					B[i][y] = SWP;
			} break;

		} fstxp (b, M);
		for (i = 1; I <= n; i++) printf ("%lld", A[ms][i));
	Putchar (' \ n ');
return 0; }

Word Explanation:

Facer: Name, Fesse

Pet:n, Pets

Give birth to:vt, having a baby.

A brood of: a Nest, a gang

SUPERVISE:VT, supervision, guidance

Peanut:n, Peanut.

PERFORM:VT, execution, completion, performance

Come up WITH:VT, raised, came up

Embarrassing:adj, embarrassing.

Quantity:n, amount, quantity