簡(jiǎn)述

對(duì)于任意情況下的蝶形求和，本質(zhì)上和任意的樹(shù)形求和是一樣的。只需要做邏輯上的抽象，假設(shè)那些點(diǎn)存在補(bǔ)全就好了，然后再加一個(gè)限定條件就ok了

代碼

#include<stdio.h> #include<string.h> #include<mpi.h> #pragma warning(disable : 4996) #define MAX_STRING 100 using namespace std; #include <fstream> #include <iostream>int main(void) {int len;double scalar, local_sum = 0;int comm_sz;int my_rank;int divided_len;int last_divided;int begin_i;int oper_N = 2;MPI_Init(NULL, NULL);MPI_Comm_size(MPI_COMM_WORLD, &comm_sz);MPI_Comm_rank(MPI_COMM_WORLD, &my_rank);// 只有一個(gè)線程的時(shí)候不操作if (comm_sz <= 1) {MPI_Finalize();return 0;}ifstream cin("D:\\C++\\VS\\repo\\MPI-DEMO\\MPI-DEMO\\input.txt");cin >> len; // 輸入數(shù)據(jù)長(zhǎng)度divided_len = len / comm_sz;last_divided = len % comm_sz;if (my_rank < last_divided) {divided_len++;begin_i = my_rank * divided_len;}else {begin_i = (my_rank - last_divided) * divided_len + ((divided_len + 1)* last_divided);}// 局部和for (int i = 0; i < len; ++i) {cin >> scalar;if (i < begin_i || i >= begin_i + divided_len) continue;else {local_sum += scalar;}}// 假設(shè)comm_sz不一定是2的冪// 蝶形求和while (oper_N <= comm_sz || (oper_N > comm_sz && oper_N / 2 < comm_sz)) {// 往前發(fā)if (my_rank % oper_N < oper_N / 2) {if (my_rank + oper_N / 2 < comm_sz){MPI_Send(&local_sum, 1, MPI_DOUBLE, my_rank + oper_N / 2, 0, MPI_COMM_WORLD);MPI_Recv(&scalar, 1, MPI_DOUBLE, my_rank + oper_N / 2, 0, MPI_COMM_WORLD, MPI_STATUSES_IGNORE);local_sum += scalar;}}// 往后發(fā)else if (my_rank % oper_N >= oper_N / 2) {MPI_Send(&local_sum, 1, MPI_DOUBLE, my_rank - oper_N / 2, 0, MPI_COMM_WORLD);MPI_Recv(&scalar, 1, MPI_DOUBLE, my_rank - oper_N / 2, 0, MPI_COMM_WORLD, MPI_STATUSES_IGNORE);local_sum += scalar;}oper_N *= 2;}if (my_rank == 0) {cout << local_sum << endl;}MPI_Finalize();return 0; }

總結(jié)

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