81 lines
2.8 KiB
C++
81 lines
2.8 KiB
C++
#include <cstdio>
|
||
#include <mpi.h>
|
||
#include <cmath>
|
||
|
||
// 计算每个进程的实际计算成本（考虑素数检测的复杂度）
|
||
long long estimate_cost(int start, int end, int step) {
|
||
long long total_cost = 0;
|
||
for (int i = start; i <= end; i += step) {
|
||
// 素数检测的成本约为 O(i)，即需要检查 i-2 次
|
||
total_cost += (i - 2);
|
||
}
|
||
return total_cost;
|
||
}
|
||
|
||
int main(int argc, char *argv[]) {
|
||
int id, p;
|
||
MPI_Init(&argc, &argv);
|
||
MPI_Comm_size(MPI_COMM_WORLD, &p);
|
||
MPI_Comm_rank(MPI_COMM_WORLD, &id);
|
||
|
||
int n = 100000;
|
||
if (argc == 2) {
|
||
n = atoi(argv[1]);
|
||
}
|
||
|
||
// 计算每个进程的计算成本
|
||
int start = 2 + id;
|
||
int end = n;
|
||
long long my_cost = estimate_cost(start, end, p);
|
||
|
||
// 收集所有进程的成本
|
||
long long *costs = nullptr;
|
||
if (id == 0) {
|
||
costs = new long long[p];
|
||
}
|
||
MPI_Gather(&my_cost, 1, MPI_LONG_LONG_INT, costs, 1, MPI_LONG_LONG_INT, 0, MPI_COMM_WORLD);
|
||
|
||
if (id == 0) {
|
||
printf("\n=== 计算成本分析 (N=%d, P=%d) ===\n", n, p);
|
||
printf("进程号\t数字数量\t估计计算成本\t成本占比\n");
|
||
printf("------------------------------------------------------------\n");
|
||
|
||
long long total_cost = 0;
|
||
for (int i = 0; i < p; i++) {
|
||
total_cost += costs[i];
|
||
}
|
||
|
||
for (int i = 0; i < p; i++) {
|
||
int count = (n - (2 + i)) / p + 1;
|
||
double percentage = 100.0 * costs[i] / total_cost;
|
||
printf("%d\t%d\t\t%lld\t\t%.2f%%\n", i, count, costs[i], percentage);
|
||
}
|
||
|
||
printf("------------------------------------------------------------\n");
|
||
printf("总计算成本: %lld\n", total_cost);
|
||
printf("平均成本: %lld\n", total_cost / p);
|
||
printf("最大成本: %lld (进程0)\n", costs[0]);
|
||
printf("最小成本: %lld (进程%d)\n", costs[p-1], p-1);
|
||
printf("\n");
|
||
|
||
double imbalance = 100.0 * (costs[0] - costs[p-1]) / (double)costs[0];
|
||
printf("=== 负载不均衡分析 ===\n");
|
||
printf("成本不均衡度: %.2f%%\n", imbalance);
|
||
printf("\n");
|
||
printf("说明：\n");
|
||
printf("- 进程0检测的数字最小（2, %d, %d, ...），但每个数字的检测成本高\n", 2+p, 2+2*p);
|
||
printf("- 进程%d检测的数字最大（%d, %d, ...），但每个数字的检测成本更高！\n", p-1, 2+(p-1), 2+2*(p-1));
|
||
printf("\n");
|
||
printf("关键问题：\n");
|
||
printf("虽然各进程检查的数字数量相近，但大数字的素数检测需要更多除法运算。\n");
|
||
printf("例如：检测2需要0次除法，检测100000需要99998次除法！\n");
|
||
printf("这导致进程间存在严重的负载不均衡。\n");
|
||
printf("\n");
|
||
|
||
delete[] costs;
|
||
}
|
||
|
||
MPI_Finalize();
|
||
return 0;
|
||
}
|