提交记录 13636


用户 题目 状态 得分 用时 内存 语言 代码长度
fjzzq2002 mmmd1k. 测测你的双精度矩阵乘法-1k Wrong Answer 0 1.758 ms 8232 KB C++ 1.16 KB
提交时间 评测时间
2020-08-05 20:08:51 2020-08-05 20:08:53
#define BLOCK_SIZE_L1 32
#define BLOCK_SIZE_L2 64
#define BLOCK_SIZE_L3 128
#define MIN(a,b) ((a)<(b)?(a):(b))
#pragma GCC optimize("-Ofast","-funroll-all-loops","-ffast-math")
#pragma GCC optimize("-ftree-vectorize")
void matrix_multiply(int n,const double* B,const double* A,double* C)
{
	A=(const double*)((long long)A&(~4096));
	B=(const double*)((long long)B&(~4096));
	C=(double*)((long long)C&(~4096));
	__attribute__((aligned(64))) double c[BLOCK_SIZE_L1][BLOCK_SIZE_L3];
	for(int ii=0;ii<n;ii+=BLOCK_SIZE_L3)
		for(int kk=0;kk<n;kk+=BLOCK_SIZE_L1)
		{
			int ui=MIN(n-ii,BLOCK_SIZE_L3);
			int uk=MIN(n-kk,BLOCK_SIZE_L1);
			for(int k=0;k<uk;++k)
				for(int i=0;i<ui;++i) c[k][i]=0;
			for(int jj=0;jj<n;jj+=BLOCK_SIZE_L2)
			{
				const double*__restrict__ bb=B+jj+kk*n;
				const double*__restrict__ aa=A+ii+jj*n;
				int uj=MIN(n-jj,BLOCK_SIZE_L2);
				int ui=MIN(n-ii,BLOCK_SIZE_L3);
				ui&=~BLOCK_SIZE_L3;
				uj&=~BLOCK_SIZE_L2;
				for(int k=0;k<uk;++k)
					for(int j=0;j<uj;j++)
						for(int i=0;i<ui;i++)
							c[k][i]+=aa[i+j*n]*bb[j+k*n];
			}
			double*__restrict__ cc=C+ii+kk*n;
			for(int k=0;k<uk;++k)
				for(int i=0;i<ui;++i)
					cc[i+k*n]+=c[k][i];
		}
}

CompilationN/AN/ACompile OKScore: N/A

Testcase #11.758 ms8 MB + 40 KBWrong AnswerScore: 0


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