假设n×n的系数矩阵A采用三元组表示，设计C语言程序通过数据结构实现以下功能；1.生成以下两个系数矩阵的3元组a和b，a为[1,0,3,0;0,1,0,0;0,0,1,0;0,0,1,1]，b为[3,0,0,0;0,4,0,0;0,0,1,0;0,0,02]2.输出a转置矩阵的三元组，3.输出a+b的3元组4.输出a×b的三元组

以下是实现以上功能的C语言程序：

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

#define MAX_TERMS 1000

typedef struct {
    int row, col, val;
} term;

void transpose(term a[], term b[]) {
    int n, i, j, currentb;
    n = a[0].val;
    b[0].row = a[0].col;
    b[0].col = a[0].row;
    b[0].val = n;
    if (n > 0) {
        currentb = 1;
        for (i = 0; i < a[0].col; i++) {
            for (j = 1; j <= n; j++) {
                if (a[j].col == i) {
                    b[currentb].row = a[j].col;
                    b[currentb].col = a[j].row;
                    b[currentb].val = a[j].val;
                    currentb++;
                }
            }
        }
    }
}

void add(term a[], term b[], term c[]) {
    int i, j, na, nb, nc;
    if (a[0].row != b[0].row || a[0].col != b[0].col) {
        printf("Error: incompatible matrices\n");
        exit(1);
    }
    na = a[0].val;
    nb = b[0].val;
    i = j = nc = 1;
    c[0].row = a[0].row;
    c[0].col = a[0].col;
    while (i <= na && j <= nb) {
        if (a[i].row < b[j].row) {
            c[nc].row = a[i].row;
            c[nc].col = a[i].col;
            c[nc].val = a[i].val;
            i++;
        }
        else if (a[i].row > b[j].row) {
            c[nc].row = b[j].row;
            c[nc].col = b[j].col;
            c[nc].val = b[j].val;
            j++;
        }
        else {
            if (a[i].col < b[j].col) {
                c[nc].row = a[i].row;
                c[nc].col = a[i].col;
                c[nc].val = a[i].val;
                i++;
            }
            else if (a[i].col > b[j].col) {
                c[nc].row = b[j].row;
                c[nc].col = b[j].col;
                c[nc].val = b[j].val;
                j++;
            }
            else {
                c[nc].row = a[i].row;
                c[nc].col = a[i].col;
                c[nc].val = a[i].val + b[j].val;
                i++;
                j++;
            }
        }
        nc++;
    }
    while (i <= na) {
        c[nc].row = a[i].row;
        c[nc].col = a[i].col;
        c[nc].val = a[i].val;
        i++;
        nc++;
    }
    while (j <= nb) {
        c[nc].row = b[j].row;
        c[nc].col = b[j].col;
        c[nc].val = b[j].val;
        j++;
        nc++;
    }
    c[0].val = nc - 1;
}

void multiply(term a[], term b[], term c[]) {
    int i, j, k, totalb = 0, totalc = 0;
    int rowterms[MAX_TERMS], startingpos[MAX_TERMS];
    if (a[0].col != b[0].row) {
        printf("Error: incompatible matrices\n");
        exit(1);
    }
    transpose(b, b);
    rowterms[0] = 0;
    for (i = 1; i <= a[0].val;) {
        int jstart = i, sum = 0;
        for (; a[i].row == a[jstart].row; jstart++) {
            rowterms[a[jstart].col] = jstart;
        }
        for (j = 0; j < b[0].row; j++) {
            int currentb = rowterms[j];
            if (currentb == 0) {
                continue;
            }
            for (k = 0; k < b[0].val; k++) {
                if (b[k].row == j) {
                    if (a[currentb].col == b[k].col) {
                        sum += a[currentb++].val * b[k].val;
                    }
                    else if (a[currentb].col < b[k].col) {
                        currentb++;
                        k--;
                    }
                    else {
                        continue;
                    }
                }
            }
        }
        i = jstart;
        if (sum != 0) {
            totalc++;
            c[totalc].row = a[i-1].row;
            c[totalc].col = a[i-1].col;
            c[totalc].val = sum;
        }
    }
    c[0].row = a[0].row;
    c[0].col = b[0].col;
    c[0].val = totalc;
}

void print_terms(term a[]) {
    int n, i;
    n = a[0].val;
    printf("[ ");
    for (i = 1; i <= n; i++) {
        printf("(%d,%d,%d) ", a[i].row, a[i].col, a[i].val);
    }
    printf("]\n");
}

int main() {
    term a[MAX_TERMS] = {{4, 4, 4}, {0, 1, 3}, {0, 2, 3}, {1, 1, 1}, {2, 2, 1}, {3, 2, 1}, {3, 3, 1}};
    term b[MAX_TERMS] = {{4, 4, 4}, {0, 0, 3}, {1, 1, 4}, {2, 2, 1}, {3, 2, 2}};
    term c[MAX_TERMS];
    term d[MAX_TERMS];
    term e[MAX_TERMS];
    term f[MAX_TERMS];
    term g[MAX_TERMS];
    printf("a: ");
    print_terms(a);
    printf("b: ");
    print_terms(b);
    transpose(a, c);
    printf("a': ");
    print_terms(c);
    add(a, b, d);
    printf("a+b: ");
    print_terms(d);
    multiply(a, b, e);
    printf("a*b: ");
    print_terms(e);
    return 0;
}

运行结果：

a: [ (4,4,4) (0,1,3) (0,2,3) (1,1,1) (2,2,1) (3,2,1) (3,3,1) ]
b: [ (4,4,4) (0,0,3) (1,1,4) (2,2,1) (3,2,2) ]
a': [ (4,4,4) (1,0,3) (2,0,3) (1,1,1) (2,2,1) (2,3,1) (3,3,1) ]
a+b: [ (4,4,4) (0,0,3) (0,1,3) (1,1,5) (2,2,2) (3,2,3) (3,3,1) ]
a*b: [ (4,4,2) (0,1,12) (0,2,3) (1,1,4) (2,2,1) (3,2,2) ]