批改娘 10106. Multiple Device (CUDA)

contents

  1. 1. 題目描述
    1. 1.1. sequence.c
  2. 2. 輸入格式
  3. 3. 輸出格式
  4. 4. 編譯參數
  5. 5. Sample Input
  6. 6. Sample Output
  7. 7. Solution

題目描述

小明的數學作業要計算方陣,現在請你幫幫他!

題目給定數個 $N \times N$ 的矩陣和 $2$ 小題。

  • $X = AB+CD$
  • $Y = ABE+CDF$

sequence.c

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#include <stdio.h>
#include <stdint.h>
// #define DEBUG
#define UINT uint32_t
#define MAXN 1024
void multiply(int N, UINT A[][MAXN], UINT B[][MAXN], UINT C[][MAXN]) {
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
UINT sum = 0; // overflow, let it go.
for (int k = 0; k < N; k++)
sum += A[i][k] * B[k][j];
C[i][j] = sum;
}
}
}
void add(int N, UINT A[][MAXN], UINT B[][MAXN], UINT C[][MAXN]) {
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++)
C[i][j] = A[i][j] + B[i][j];
}
}
void rand_gen(UINT c, int N, UINT A[][MAXN]) {
UINT x = 2, n = N*N;
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
x = (x * x + c + i + j)%n;
A[i][j] = x;
}
}
}
void print_matrix(int N, UINT A[][MAXN]) {
for (int i = 0; i < N; i++) {
fprintf(stderr, "[");
for (int j = 0; j < N; j++)
fprintf(stderr, " %u", A[i][j]);
fprintf(stderr, " ]\n");
}
}
UINT signature(int N, UINT A[][MAXN]) {
UINT h = 0;
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++)
h = (h + A[i][j]) * 2654435761LU;
}
return h;
}
UINT IN[6][MAXN][MAXN], TMP[6][MAXN][MAXN];
int main() {
int N, S[6];
scanf("%d", &N);
for (int i = 0; i < 6; i++) {
scanf("%d", &S[i]);
rand_gen(S[i], N, IN[i]);
}
// AB
multiply(N, IN[0], IN[1], TMP[0]);
// CD
multiply(N, IN[2], IN[3], TMP[1]);
// AB+CD
add(N, TMP[0], TMP[1], TMP[2]);
printf("%u\n", signature(N, TMP[2]));

// ABE
multiply(N, TMP[0], IN[4], TMP[3]);
// CDF
multiply(N, TMP[1], IN[5], TMP[4]);
// ABE+CDF
add(N, TMP[3], TMP[4], TMP[5]);
printf("%u\n", signature(N, TMP[5]));
return 0;
}

輸入格式

輸入有多組測資,每組第一行會有一個整數 $N$,表示題目給定 $N \times N$ 矩陣,第二行上會有 $6$ 個整數,分別為矩陣 $A, B, C, D, E, F$ 的生成種子。

  • $1 \le N \le 1024$
  • $0 \le S_i \le 2^{31}$

輸出格式

輸出兩行 $X$ 和 $Y$ 的雜湊值,可參考 sequence.c 的流程。

編譯參數

1
$ nvcc -Xcompiler "-O2 -fopenmp" main.cu -o main

Sample Input

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0 1 2 3 4 5
10
0 1 2 3 4 5

Sample Output

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2385860290
1374821695
617438354
1897844131

Solution

與 OpenMP 一起設計,仿造 OpenCL 版本。

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#include <stdio.h>
#include <assert.h>
#include <inttypes.h>
#include <string.h>
#include <cuda.h>

#define MAXGPU 3
#define MAXN 1024
#define GPULOCAL 64
#define UNLOOP 8
#define CheckErr(status) { gpuAssert((status), __FILE__, __LINE__); }
inline void gpuAssert(cudaError_t code, const char *file, int line, int abort=true) {
if (code != cudaSuccess) {
fprintf(stderr,"GPUassert: %s %s %d\n", cudaGetErrorString(code), file, line);
if (abort) exit(code);
}
}
uint32_t hostMtx[MAXGPU][6][MAXN*MAXN];
uint32_t hostMid[MAXGPU][2][MAXN*MAXN];

__global__ void matrixMul(uint32_t A[], uint32_t B[], uint32_t C[], int N) {
int r = blockIdx.x * blockDim.x + threadIdx.x;
int x = r / N, y = r % N;
uint32_t sum = 0;
for (int i = 0; i < N; i++)
sum += A[x*N + i] * B[i*N + y];
C[x * N + y] = sum;
}
__global__ void matrixAdd(uint32_t A[], uint32_t B[], uint32_t C[]) {
int r = blockIdx.x * blockDim.x + threadIdx.x;
C[r] = A[r] + B[r];
}


int readIn(uint32_t S[], int *n, int devId) {
int N, M;
if (scanf("%d", &N) != 1)
return 0;
M = 6;
for (int i = 0; i < M; i++)
assert(scanf("%d", &S[i]) == 1);

for (int p = 0; p < M; p++)
#pragma omp task
{
uint32_t x = 2, n = N*N, c = S[p];
x = 2;
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
x = (x * x + c + i + j)%n;
hostMtx[devId][p][i*N+j] = x;
}
}
}
#pragma omp taskwait
*n = N;
return 1;
}
uint32_t writeOut(uint32_t *hostC, int N) {
uint32_t h = 0;
uint32_t *Cend = hostC + N*N, *C = hostC;
for (; C != Cend; C++)
h = (h + *C) * 2654435761LU;
return h;
}
void matrix_mul(int N, uint32_t *cuMtxA, uint32_t *cuMtxB, uint32_t *cuMtxC) {
int localSz = 1;
for (int i = 1; i <= 1024; i++) {
if (N*N % i == 0)
localSz = i;
}
dim3 cuBlock(localSz);
dim3 cuGrid(N*N/localSz);
matrixMul<<<cuGrid, cuBlock>>>(cuMtxA, cuMtxB, cuMtxC, N);
CheckErr(cudaGetLastError());
}
void matrix_add(int N, uint32_t *cuMtxA, uint32_t *cuMtxB, uint32_t *cuMtxC) {
int localSz = 1;
for (int i = 1; i <= 1024; i++) {
if (N*N % i == 0)
localSz = i;
}
dim3 cuBlock(localSz);
dim3 cuGrid(N*N/localSz);
matrixAdd<<<cuGrid, cuBlock>>>(cuMtxA, cuMtxB, cuMtxC);
CheckErr(cudaGetLastError());
}
void solver(int devId, int N, uint32_t *cuMtx[], uint32_t *cuMtxTmp[], uint32_t ret[]) {
uint32_t memSz = N*N*sizeof(uint32_t);
for (int i = 0; i < 6; i++) {
cudaMemcpy(cuMtx[i], hostMtx[devId][i], memSz, cudaMemcpyHostToDevice);
CheckErr(cudaGetLastError());
}
// cuMtxTmp[0] = AB
matrix_mul(N, cuMtx[0], cuMtx[1], cuMtxTmp[0]);
// cuMtxTmp[1] = CD
matrix_mul(N, cuMtx[2], cuMtx[3], cuMtxTmp[1]);
// cuMtxTmp[2] = ABE
matrix_mul(N, cuMtxTmp[0], cuMtx[4], cuMtxTmp[2]);
// cuMtxTmp[3] = CDF
matrix_mul(N, cuMtxTmp[1], cuMtx[5], cuMtxTmp[3]);
// cuMtxTmp[4] = AB + CD
matrix_add(N, cuMtxTmp[0], cuMtxTmp[1], cuMtxTmp[4]);
// cuMtxTmp[5] = ABE+CDF
matrix_add(N, cuMtxTmp[2], cuMtxTmp[3], cuMtxTmp[5]);

cudaMemcpy(hostMid[devId][0], cuMtxTmp[4], memSz, cudaMemcpyDeviceToHost);
cudaMemcpy(hostMid[devId][1], cuMtxTmp[5], memSz, cudaMemcpyDeviceToHost);

for (int i = 0; i < 2; i++)
#pragma omp task
{
ret[i] = writeOut(hostMid[devId][i], N);
}
#pragma omp taskwait
}
int main(int argc, char *argv[]) {
uint32_t *cuMtx[MAXGPU][6], *cuMtxTmp[MAXGPU][6];
uint32_t memSz = MAXN*MAXN*sizeof(uint32_t);
for (int p = 0; p < MAXGPU; p++) {
cudaSetDevice(p);
for (int i = 0; i < 6; i++) {
cudaMalloc((void **) &cuMtx[p][i], memSz);
CheckErr(cudaGetLastError());
}
for (int i = 0; i < 6; i++)
cudaMalloc((void **) &cuMtxTmp[p][i], memSz);
}
int inN = 0;
static uint32_t ansQue[32767][2];
#pragma omp parallel sections
{
#pragma omp section
{
cudaSetDevice(0);
while (1) {
int f = 0, N, pid = 0;
uint32_t S[32];
#pragma omp critical
{
f = readIn(S, &N, 0);
pid = inN;
inN += f;
}
if (f == 0)
break;
solver(0, N, cuMtx[0], cuMtxTmp[0], ansQue[pid]);
}
}

#pragma omp section
{
cudaSetDevice(1);
while (1) {
int f = 0, N, pid = 0;
uint32_t S[32];
#pragma omp critical
{
f = readIn(S, &N, 1);
pid = inN;
inN += f;
}
if (f == 0)
break;
solver(1, N, cuMtx[1], cuMtxTmp[1], ansQue[pid]);
}
}
#pragma omp section
{
cudaSetDevice(2);
while (1) {
int f = 0, N, pid = 0;
uint32_t S[32];
#pragma omp critical
{
f = readIn(S, &N, 2);
pid = inN;
inN += f;
}
if (f == 0)
break;
solver(2, N, cuMtx[2], cuMtxTmp[2], ansQue[pid]);
}
}

}
for (int i = 0; i < inN; i++)
printf("%u\n%u\n", ansQue[i][0], ansQue[i][1]);
for (int i = 0; i < 6; i++)
cudaFree(cuMtx[i]);
for (int i = 0; i < 6; i++)
cudaFree(cuMtxTmp[i]);
return 0;
}