梯形积分法串行代码的实现:
#define _CRT_SECURE_NO_WARNINGS 1
#include <stdio.h>
#include <math.h>
#define MAXN 30
double fun1(double x);
double fun2(double x);
double fun3(double x);
double Definite_integral(double a, double b, double (*p)(double x));
int main()
{
int c = 1;
double a, b;
char d = '0';
printf("*************************************\n");
printf("***** 用梯形积分串行算法 *******\n");
printf("***** 1、函数为：y = x *******\n");
printf("***** 2、函数为：y = x^2 *******\n");
printf("***** 3、函数为：y = sin(x) *******\n");
printf("*************************************\n");
while (0 != c) {
printf("请输入您要选择的类型：");
scanf("%d",&c);
switch (c)
{
case 1:
printf("请输入积分上下限(以空格隔开)：>");
scanf("%lf %lf",&a,&b);
printf("该函数在%lf-%lf上的积分值为：》%lf\n", a, b, Definite_integral(a, b, fun1));
break;
case 2:
printf("请输入积分上下限(以空格隔开)：>");
scanf("%lf %lf",&a,&b);
printf("该函数在%lf-%lf上的积分值为：》%lf\n", a, b, Definite_integral(a, b, fun1));
break;
case 3:
printf("请输入积分上下限(以空格隔开)：>");
scanf("%lf %lf",&a,&b);
printf("该函数在%lf-%lf上的积分值为：》%lf\n", a, b, Definite_integral(a, b, fun1));
break;
default:
printf("您的输入有误，您的选择只能是：1、2、3、0，请重新输入！\n");
break;
}
printf("您还要继续选择吗？（1/0）:>");
scanf("%d",&c);
}
printf("程序结束，欢迎下次再来！");
return 0;
}
/* 梯形法求定积分*/
double Definite_integral(double a, double b, double (*p)(double x)) {
int i, n = MAXN;
double x, y1, y2, deta, f;
deta = (b - a) / n;
x = a;
y1 = (*p)(x);
f = 0;
for (i = 0; i < n; i++) {
x += deta;
y2 = (*p)(x);
f += (y1 + y2) * deta / 2;
y1 = y2;
}
return f;
}
/*定义被积函数,三种供选择*/
double fun1(double x) {
double fx;
fx = x;
return fx;
}
double fun2(double x) {
double fx;
fx = x * x;
return fx;
}
double fun3(double x) {
double fx;
fx = sin(x);
return fx;
}
结果：
使用MPI点对点通信梯形积分法并行代码的实现
#define _CRT_SECURE_NO_WARNINGS 1
#include<stdio.h>
#include "mpi.h"
//输入
void Get_input(int my_rank, int comm_sz, double* a_p, double* b_p, int* n_p);
double Trap(double left_endpt, double right_endpt, int trap_count, double b);
double f(double x);
//使用MPI点对点通信梯形积分法并行代码的实现
int main(void) {
int my_rank, comm_sz, n = 1000000000, local_n;//n为分成梯形的个数，a为积分下线，b为积分上线
double a = 0.0, b = 1.0, h, local_a, local_b;
double local_int, total_int;
int source;
double* a_p = &a;
double* b_p = &b;
int* n_p = &n;
MPI_Init(NULL, NULL);
MPI_Comm_rank(MPI_COMM_WORLD, &my_rank);
MPI_Comm_size(MPI_COMM_WORLD, &comm_sz);
Get_input(my_rank, comm_sz, a_p, b_p, n_p);
h = (b - a) / n;
local_n = n / comm_sz;
local_a = a + my_rank * local_n * h;
local_b = local_a + local_n * h;
local_int = Trap(local_a, local_b, local_n, h);
if (my_rank != 0) {
MPI_Send(&local_int, 1, MPI_DOUBLE, 0, 0, MPI_COMM_WORLD);
}
else {
total_int = local_int;
for (source = 1; source < comm_sz; source++) {
MPI_Recv(&local_int, 1, MPI_DOUBLE, source, 0, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
total_int += local_int;
}
}
if (my_rank == 0) {
printf("With n = %d trapezoids, our estimate\n", n);
printf("of the integral from %f to %f = %.15e\n", a, b, total_int);
}
MPI_Finalize();
return 0;
}
//梯形面积和 = h * ( f(x0)/2 + f(x1) + ....... + f(xn)/2) h = (b-a)/n;
double Trap(double left_endpt, double right_endpt, int trap_count, double base_len) {
double estimate, x;
int i;
estimate = (f(left_endpt) + f(right_endpt)) / 2.0;
for (i = 1; i <= trap_count - 1; i++) {
x = left_endpt + i * base_len;
estimate += f(x);
}
estimate = estimate * base_len;
return estimate;
}
double f(double x) {
return 4.0 / (1 + x * x);
}
void Get_input(int my_rank, int comm_sz, double* a_p, double* b_p, int* n_p) {
int dest;
if (my_rank == 0) {
printf("Enter a, b,and n \n");
scanf("%lf %lf %d", a_p, b_p, n_p);
for (dest = 1; dest < comm_sz; dest++) {
MPI_Send(a_p, 1, MPI_DOUBLE, dest, 0, MPI_COMM_WORLD);
MPI_Send(b_p, 1, MPI_DOUBLE, dest, 0, MPI_COMM_WORLD);
MPI_Send(n_p, 1, MPI_INT, dest, 0, MPI_COMM_WORLD);
}
}
else {
MPI_Recv(a_p, 1, MPI_DOUBLE, 0, 0, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
MPI_Recv(b_p, 1, MPI_DOUBLE, 0, 0, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
MPI_Recv(n_p, 1, MPI_INT, 0, 0, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
}
}
结果
使用MPI集合通信梯形积分法并行代码的实现。
#define _CRT_SECURE_NO_WARNINGS 1
#include<stdio.h>
#include "mpi.h"
double Trap(double left_endpt, double right_endpt, int trap_count, double b);
double f(double x);
void Get_input(int my_rank, int comm_sz, double* a_p, double* b_p, int* n_p);
int main(void) {
int my_rank, comm_sz, n, local_n;
double a, b, h, local_a, local_b;
double local_int, total_int;
int source;
MPI_Init(NULL, NULL);
MPI_Comm_rank(MPI_COMM_WORLD, &my_rank);
MPI_Comm_size(MPI_COMM_WORLD, &comm_sz);
Get_input(my_rank, comm_sz, &a, &b, &n);
h = (b - a) / n;
local_n = n / comm_sz;
local_a = a + my_rank * local_n * h;
local_b = local_a + local_n * h;
local_int = Trap(local_a, local_b, local_n, h);
MPI_Reduce(&local_int, &total_int, 1, MPI_DOUBLE, MPI_SUM, 0, MPI_COMM_WORLD);
if (my_rank == 0) {
printf("With n = %d trapezoids, our estimate\n", n);
printf("of the integral from %f to %f = %.15e\n", a, b, total_int);
}
MPI_Finalize();
return 0;
}
double Trap(double left_endpt, double right_endpt, int trap_count, double base_len) {
double estimate, x;
int i;
estimate = (f(left_endpt) + f(right_endpt)) / 2.0;
for (i = 1; i <= trap_count - 1; i++) {
x = left_endpt + i * base_len;
estimate += f(x);
}
estimate = estimate * base_len;
return estimate;
}
double f(double x) {
return 4.0 / (1 + x * x);
}
void Get_input(int my_rank, int comm_sz, double* a_p, double* b_p, int* n_p) {
int dest;
if (my_rank == 0) {
printf("Enter a, b, and n\n");
scanf("%lf %lf %d", a_p, b_p, n_p);
}
MPI_Bcast(a_p, 1, MPI_DOUBLE, 0, MPI_COMM_WORLD);
MPI_Bcast(b_p, 1, MPI_DOUBLE, 0, MPI_COMM_WORLD);
MPI_Bcast(n_p, 1, MPI_INT, 0, MPI_COMM_WORLD);
}
结果：
