upstream: add GEMM kernel references from 4 repos for BI-V100 porting

Sources (all CUDA 10.2 compatible, no CUTLASS/Triton dependency):
- leimao/CUDA-GEMM-Optimization: v00-v07, fp16 WMMA variant, double buffered
- siboehm/SGEMM_CUDA: kernel 1-12, warp tiling + double buffering
- wangzyon/NVIDIA_SGEMM_PRACTICE: kernel 1-7
- edtallison/sgemm-cuda: kernel 1-12 (reimplementation with notes)

Key porting issue: ALL kernels hardcode WARPSIZE=32.
BI-V100 has warp_size=64. Need to:
1. Replace all 32U / WARPSIZE constants with 64
2. Adjust warp subtile decomposition (WMITER, WNITER, WSUBM, WSUBN)
3. Adjust shared memory bank conflict avoidance (may have different bank count)
4. Test __shfl_down_sync with mask=0xFFFFFFFFFFFFFFFF (64-bit)

upstream_ref/nvidia_sgemm_practice/kernel_3.cuh

@@ -0,0 +1,71 @@
+#pragma once
+
+#include <cuda_runtime.h>
+#include <cublas_v2.h>
+#include <stdio.h>
+#include <stdlib.h>
+
+template<const int BM,
+        const int BN,
+        const int BK,
+        const int TM>
+__global__ void mysgemm_v3(int M, int N, int K, float alpha, float *A, float *B, float beta, float *C) {
+    int bx = blockIdx.x;
+    int by = blockIdx.y;
+    int thread_num = BM * BN / TM; // 一个线程负责block中计算TM个元素
+
+    int tx = threadIdx.x % BN;
+    int ty = threadIdx.x / BN * TM;
+
+    __shared__ float As[BM * BK];
+    __shared__ float Bs[BK * BN];
+
+    // 移动到当前block
+    A = &A[by * BM * K];
+    B = &B[bx * BN];
+    C = &C[by * BM * N + bx * BN];
+
+    /*
+    当前线程负责搬运全局内存中第a_tile_row行，第a_tile_col列元素至共享内存第a_tile_row行，第a_tile_col列
+    a_tile_stride表示block中线程可搬运a_tile_stride行至共享内存；
+
+    若BM=64,BK=8,thread_num=512,则a_tile_stride=64,a_tile_stride=BM，表示每个线程搬运一轮即可完成所需元素的搬运;
+    若BM=128,BK=8,thread_num=512,则a_tile_stride=64,表示每个线程搬运两轮即可完成所需元素的搬运;
+    */
+    int a_tile_row = threadIdx.x / BK;
+    int a_tile_col = threadIdx.x % BK;
+    int a_tile_stride = thread_num / BK;
+
+    int b_tile_row = threadIdx.x / BN;
+    int b_tile_col = threadIdx.x % BN;
+    int b_tile_stride = thread_num / BN;
+
+    float tmp[TM + 1] = {0.}; // 每个线程负责TM个元素，则需要申请TM个寄存器保存累加值，额外的一个寄存器用于缓存；
+    #pragma unroll
+    for (int k = 0; k < K; k += BK) {
+        #pragma unroll
+        for (int i = 0; i < BM; i += a_tile_stride) {
+            As[(a_tile_row + i) * BK + a_tile_col] = A[(a_tile_row + i) * K + a_tile_col];
+        }
+        #pragma unroll
+        for (int i = 0; i < BK; i += b_tile_stride) {
+            Bs[(b_tile_row + i) * BN + b_tile_col] = B[(b_tile_row + i) * N + b_tile_col];
+        }
+        __syncthreads();
+        A += BK;
+        B += BK * N;
+        #pragma unroll
+        for (int i = 0; i < BK; i++) {
+            tmp[TM] = Bs[tx + i * BN]; // 额外的一个寄存器，避免反复从共享内存中读取Bs[tx + i * BN]
+            #pragma unroll  // 循环展开，增加指令并行度
+            for (int j = 0; j < TM; j++) {
+                tmp[j] += As[(ty + j) * BK + i] * tmp[TM];
+            }
+        }
+        __syncthreads();
+    }
+    #pragma unroll
+    for (int j = 0; j < TM; j++) {
+        C[(ty + j) * N + tx] = alpha * tmp[j] + beta * C[(ty + j) * N + tx];
+    }
+}