集合上机实验

一：真值计算
主要考虑计算机如何表示逻辑演绎，且，或，非，蕴含，双条件。
从键盘输入两个命题P和Q的真值，求它们的合取、析取、条件和双条件的真值。用C语言实现。
二:集合的运算
集合的交，并，差，补，通过for,for双循环，类似命题：任意的x任意的y,都成立的一个表达式（x+y=y+x).每个x,都把所有y遍历一次。

一,用无序的办法表示集合,有非常多的检索。

#define _CRT_SECURE_NO_WARNINGS
#include
#include

//int creatjh(int *a[]);
 void Both(int a[], int i, int b[], int j,int *c);
int Isin(int a,int b[],int n);
int IN(int a[],int i, int b[],int j,int *c);
void All(int a[], int i, int b[], int j, int *c);
void equal(int a[], int i, int ib[], int j);
int cdseris(int a[], int i, int b[], int j);
int main()
{
    int jh1[10] = { 2,5,3,7,5 };
    int jh2[10] = { 5,3,9,7,6 };
    int jh3[10] = { 5,3,3,8,1,4};
    int jh4[10] = { 5,3,3,6,6,4 }; 
    int both[100] = { 0 };
    int both2[100] = { 0 };
    int both3[100] = { 0 };
    int all1[20] = { 0 };
    int all2[30] = { 0 };
    int all3[40] = { 0 };
    int e = 9;
    int n1 = 10;
    int n2=10,n3 = 10,n4=10;
    int n5 = 20;
    Isin(e, jh2, n2);
    printf("\n");
    All(jh1, n1, jh2, n2, all1);
    printf("\n");
    All(all1,n5, jh3,n3, all2);
    printf("\n");
    int cd=cdseris(jh4, n4, jh1, n1);
    printf("\n");
    equal(jh1, n1, jh4, n4);
    return 0;
}
int Isin(int c, int b[],int n) {
    int is;
    for (int i = 0; i < n+1; i++) {
        if (b[i] == c) {
            is = 1;
            break;
        }
        if (i == n) {
            is = 0;
        }
    }
    return is;
}
int IN(int a[], int i,int b[],int j,int *c) {
    int v = 0;
    for (int k = 0; k < i; k++)
    {
        int st = 0;
        for (int ww = 0; ww < v+1; ww++) 
        {
            if (c[ww] == a[k]) {
                st = 1;
                break;
            }
        }
        if (st == 1) 
        {
            continue;
        }
        for (int s = 0; s < j+1; s++) 
        {
            if (b[s] == a[k]) 
            {
                break;
            }
            if (s == j)
            {
                c[v] = a[k];
                printf("%d\t", c[v]);
                v++;
            }
        }
    }
}
int creatjh(int a[]) {
    int j = 0;
    int n = 0;
    int y;
    while (1) {
        scanf("%d", &y);
        if (y =='\0') {
            break;
        }
        a[j++] = y;
        n++;
    }
    return n;

}
void All(int a[], int i, int b[], int j, int *c) {
    int bb = i > j ? i : j;
    int v = IN(a, i, b, j, c);
    for (int s = 0; s < j; s++)
    {       
        int st = 1;
            for (int ww = 0; ww < v; ww++) 
            {
                if (c[ww]==b[s]) {
                    st = 0;
                    break;
                }
            }
            if (st == 0) {
                continue;
            }
            c[v] = b[s];
            printf("%d\t", c[v]);
            v++;
    }

}
void Both(int a[], int mi, int b[], int nj, int *c){
   int v = 0;
   int st = 0;
    int bb = mi > nj ? mi : nj;
    for (int k = 0; k < mi; k++)
    {
        for (int ww = 0; ww < v+1; ww++) {
            if (c[ww] == a[k]) {
                st = 1;
                break;
            }
        }
        if (st == 1) {
            continue;
        }
        for (int s = 0; s < nj; s++)
            if (b[s] == a[k]) {
                c[v] = a[k];
                printf("%d\t", c[v]);
                v++;
                break;
            }
    }
}
int cdseris(int a[], int i, int b[], int j) {
    int is=1;
    for (int k = 0; k < i; k++) {
        is = Isin(a[k], b, 10);
        if (is == 0) {
            printf("not cd seris\n");
            break;
        }
    }
    if (is == 1) {
        printf("yes,is cd seris\n");
    }
        return is;
}
void equal(int a[], int i, int b[], int j){
    int is = 0;
    int iss = 0;
    is = cdseris(a, i, b, j);
    iss = cdseris(b, j, a, i);

    if (is == 1 && iss == 1) {
        printf("yes,is equal");
    }
    else
        printf("no,is not equal");
}

二,用1，0表示法，可以快捷计算。

#define _CRT_SECURE_NO_WARNINGS
#include
#include
void display(int *aa, int *p);
void fei(int *aa,int *p);
void IN(int *a, int *b, int *p);
void Both(int *a, int *b, int *p);
void All(int *a, int *b, int *p);

int main()
{
    int seris[10] = { 1,2,3,4,5,6,7,8,9,10 };
    int sp[10] = { 11,22,33,44,55,66,77,88,99,100 };
    int a[10] = { 1,0,0,1,1,0,0,1,1,0 };
    int b[10] = { 0,0,1,1,0,0,1,1,0,0 };
    for (int y = 0; y < 10; y++)
    {
        int q = seris[y];
        int w = sp[y];
        int e = a[y];
        int r = b[y];
        printf("%d\t", q);
        printf("%d\t", w);
        printf("%d\t", e);
        printf("%d\t", r);
        printf("\n");
    }
    display(a,seris);//display a
    display(b,seris);
    fei(a,seris);//not a
    IN(a, b, seris);//a-b
    Both(a, b, seris);//a^b
    All(a, b, seris);


}
void display(int *aa,int *p) {
    for (int i = 0; i < 10; i++)
        if (aa[i] == 1) {
            printf("%d", p[i]);
        }
    printf("\n");
}
void fei(int *a,int *p) {
    int af[10] = { 0 };
    for (int i = 0; i < 10; i++) 
    {
        if (a[i] == 0) {
            af[i] = 1;
        }
    }
    for (int j = 0; j < 10; j++) 
    {
        printf("%d", af[j]);
    }
    printf("\t");
    display(af, p);
}
void IN(int *a, int *b,int *p) {
    int a_b[10] = { 0 };
    for (int i = 0; i < 10; i++)
        if (a[i] == 1 && b[i] == 0) {
            a_b[i] = 1;
        }
    for (int j = 0; j < 10; j++) {
        printf("%d", a_b[j]);
    }
    printf("\t");
    display(a_b, p);
}
void Both(int *a, int *b, int *p)
{
    int ajb[10] = { 0 };
    for (int i = 0; i < 10; i++)
        if (a[i] == 1 && b[i] == 1) {
            ajb[i] = 1;
        }
    for (int j = 0; j < 10; j++) {
        printf("%d", ajb[j]);
    }
    printf("\t");
    display(ajb, p);
}
void All(int *a, int *b, int *p) {
    int ab[10] = { 0 };
    for (int i = 0; i < 10; i++)
        if (a[i] == 1 || b[i] == 1) {
            ab[i] = 1;
        }
    for (int j = 0; j < 10; j++) {
        printf("%d", ab[j]);
    }
    printf("\t");
    display(ab, p);
}

这个时候，如何进行组合的计算呢？