并行计算上机代码

Posted 2021-02-13 solvit

多线程求PI：

#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>

double res;

typedef struct Args {
    long n;
    long begin;
    long step;
} Args;

void *cal(void *arg);

int main(int argc, char *argv[]) {
    long i;
    long n = atol(argv[1]);
    int np = atoi(argv[2]);
    Args *arg;
    double* val;

    pthread_t *pid;
    pid = (pthread_t *)malloc(np * sizeof(pthread_t));
    
    res = 0.0;
    for (i = 0; i < np; i++) {
        arg = (Args *)malloc(sizeof(Args));
        arg->n = n;
        arg->begin = 2*i+1;
        arg->step = np;
        pthread_create(&pid[i], NULL, cal, (void *)arg);
    }
    for (i = 0; i < np; i++) {
        pthread_join(pid[i], (void *)val);
    }
    res = res * 4.0;
    printf("%lf
", res);
    free(pid);
    return 0;
}

void *cal(void *_arg) {
    long i;
    double val;
    Args *arg = (Args *)_arg;
    
    val = 0.0;
    for (i = arg->begin; i <= arg->n; i += 2*arg->step) {
        if(i % 4 == 3) val -= 1.0 / i;
        else val += 1.0 / i;
    }
    res += val;
    return NULL;
}

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多线程矩阵乘法：

#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>

#define maxn 2

typedef struct Args {
    long n;
    long begin;
    long step;
} Args;

typedef struct Matrix{
    double a[maxn][maxn];
    /*
    void init(){
        for(int i=0;i<maxn;i++){
            for(int j=0;j<maxn;j++){
                a[i][j]=0.0;
            }
        }
        for(int i=0;i<maxn;i++){
            a[i][i] = 1.0;
        }
    }*/
} Matrix;

Matrix res, A;

void *cal(void *arg);
void output(Matrix a);
Matrix mul(Matrix a, Matrix b);
Matrix qpow(Matrix a, int n);

int main(int argc, char *argv[]) {
    long i;
    long n = atol(argv[1]);
    int np = atoi(argv[2]);
    Args *arg;

    pthread_t *pid;
    pid = (pthread_t *)malloc(np * sizeof(pthread_t));

    //res.init();
    for(int i=0;i<maxn;i++){
        for(int j=0;j<maxn;j++){
            res.a[i][j]=0.0;
        }
    }
    for(int i=0;i<maxn;i++){
        res.a[i][i] = 1.0;
    }
    //
    for(int i=0;i<maxn;i++){
        for(int j=0;j<maxn;j++){
            scanf("%lf", &A.a[i][j]);
        }
    }
    for (i = 0; i < np; i++) {
        arg = (Args *)malloc(sizeof(Args));
        arg->n = n;
        arg->begin = i+1;
        arg->step = np;
        pthread_create(&pid[i], NULL, cal, (void *)arg);
    }
    for (i = 0; i < np; i++) {
        pthread_join(pid[i], NULL);
    }
    output(res);
    free(pid);
    return 0;
}

void *cal(void *_arg) {
    long i;
    Matrix val;
    Args *arg = (Args *)_arg;

    //val.init();
    for(int i=0;i<maxn;i++){
        for(int j=0;j<maxn;j++){
            val.a[i][j]=0.0;
        }
    }
    for(int i=0;i<maxn;i++){
        val.a[i][i] = 1.0;
    }
    for (i = arg->begin; i <= arg->n; i += arg->step) {
        val = mul(val, A);
    }
    res = mul(res, val);
    return NULL;
}

Matrix mul(Matrix a, Matrix b){
    Matrix ans;
    for(int i=0;i<maxn;i++){
        for(int j=0;j<maxn;j++){
            ans.a[i][j] = 0;
            for(int k=0;k<maxn;k++){
                ans.a[i][j] += a.a[i][k] * b.a[k][j];
            }
        }
    }
    return ans;
}

Matrix qpow(Matrix a, int n){
    Matrix ans;
    //ans.init();
    for(int i=0;i<maxn;i++){
        for(int j=0;j<maxn;j++){
            ans.a[i][j]=0.0;
        }
    }
    for(int i=0;i<maxn;i++){
        ans.a[i][i] = 1.0;
    }
    while(n > 0){
        if(n&1) ans = mul(ans, a);
        a = mul(a, a);
        n >>= 1;
    }
    return ans;
}

void output(Matrix a){
    for(int i=0;i<maxn;++i){
        for(int j=0;j<maxn;++j){
            printf("%lf ", a.a[i][j]);
        }
        puts("");
    }
}

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