Write a parallel version of the program below using MPI. The output should be essentially the same as the sequential version; the only exception is that the perfect numbers might be found in a different order. Your program should report the total number of perfect numbers found and the time, exactly as the sequential version does. #include #include #include double mytime() { struct timeval t; gettimeofday(&t, NULL); return t.tv_sec + t.tv_usec/1000000.0; } _Bool is_perfect(int n) { if (n < 2) return 0; int sum = 1, i = 2; while (1) { const int i_squared = i*i; if (i_squared < n) { if (n%i == 0) sum += i + n/i; i++; } else { if (i_squared == n) sum += i; break; } } return sum == n; } int main(int argc, char * argv[]) { int num_perfect = 0; double start_time = mytime(); if (argc != 2) { printf("Usage: perfect.exec bound\n"); exit(1); } int bound = atoi(argv[1]); for (int i=1; i<=bound; i++) { if (i%1000000 == 0) { printf("i = %d\n", i); fflush(stdout); } if (is_perfect(i)) { printf("Found a perfect number: %d\n", i); num_perfect++; } } printf("Number of perfect numbers less than or equal to %d: %d. Time = %lf\n\ ", bound, num_perfect, mytime() - start_time); }
Write a parallel version of the program below using MPI. The output should be essentially the same as the sequential version; the only exception is that the perfect numbers might be found in a different order. Your program should report the total number of perfect numbers found and the time, exactly as the sequential version does.
#include <stdio.h>
#include <stdlib.h>
#include <sys/time.h>
double mytime() {
struct timeval t;
gettimeofday(&t, NULL);
return t.tv_sec + t.tv_usec/1000000.0;
}
_Bool is_perfect(int n) {
if (n < 2) return 0;
int sum = 1, i = 2;
while (1) {
const int i_squared = i*i;
if (i_squared < n) {
if (n%i == 0) sum += i + n/i;
i++;
} else {
if (i_squared == n) sum += i;
break;
}
}
return sum == n;
}
int main(int argc, char * argv[]) {
int num_perfect = 0;
double start_time = mytime();
if (argc != 2) {
printf("Usage: perfect.exec bound\n");
exit(1);
}
int bound = atoi(argv[1]);
for (int i=1; i<=bound; i++) {
if (i%1000000 == 0) {
printf("i = %d\n", i);
fflush(stdout);
}
if (is_perfect(i)) {
printf("Found a perfect number: %d\n", i);
num_perfect++;
}
}
printf("Number of perfect numbers less than or equal to %d: %d. Time = %lf\n\
",
bound, num_perfect, mytime() - start_time);
}
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