1.
编写一个算法，计算n阶方阵A的两个对角线元素的和。A采用数组表示。
#include <stdio.h>#define N 4int main(){ int A[N][N]; int i,j,m; for(i=0;i<N;i++){ for(j=0;j<N;j++){ scanf("%d",&m); A[i][j]=m; } }/* for(i=0;i<N;i++){ for(j=0;j<N;j++){ printf("%4d",A[i][j]); } printf("%d
"); }*/ //计算主对角线的和 int sum1=0,sum2=0; for(i=0;i<N;i++){ sum1+=A[i][i]; } //计算副对角线的和 for(i=0;i<N;i++){ sum2+=A[i][N-i-1]; } printf("The sum of the main diagonals is %d
the sum of the secondary diagonals is %d",sum1,sum2);} return 0;2.
编写一个算法，实现两个采用三元组表示的n阶稀疏矩阵的加法运算。
#include <stdio.h>#include <stdlib.h>#include <time.h>#define M 5#define N 6struct part{ int row; int col; int value;};int main(){ int A[M][N] ={0}; int B[M][N] ={0}; int C[M][N]; srand(time(NULL)); int n =1 + rand()%(M*N/3); //大致保证非零数在1到10之间 struct part data1[n],data2[n],data3[2*n]; printf("n = %d
",n); int i,j; for(i=0;i<n;i++){ A[rand()%M][rand()%N] = 1+rand()%10; B[rand()%M][rand()%N] = 1+rand()%10; } int a=0,b=0,c=0,k=0; for(i=0;i<M;i++){ for(j=0;j<N;j++){ if(A[i][j]!=0){ data1[a].row = i; data1[a].col = j; data1[a].value = A[i][j]; a++; } printf("%4d",A[i][j]); } printf("
"); } for(i=0;i<n;i++){ printf("(%d %d %d)
",data1[i].row,data1[i].col,data1[i].value); } printf("
"); for(i=0;i<M;i++){ for(j=0;j<N;j++){ if(B[i][j]!=0){ data2[b].row = i; data2[b].col = j; data2[b].value = B[i][j]; b++; } printf("%4d",B[i][j]); } printf("
"); } for(i=0;i<n;i++){ printf("(%d %d %d)
",data2[i].row,data2[i].col,data2[i].value); } for(i=0;i<M;i++){ for(j=0;j<N;j++){ C[i][j] = A[i][j]+B[i][j]; if((A[i][j]!=0)&&(B[i][j]!=0)){ k++; } } } for(i=0;i<M;i++){ for(j=0;j<N;j++){ if(C[i][j]!=0){ data3[c].row = i; data3[c].col = j; data3[c].value = C[i][j]; c++; } printf("%4d",C[i][j]); } printf("
"); } printf("k = %d
",k); for(i=0;i<2*n-k;i++){ printf("(%d %d %d)
",data3[i].row,data3[i].col,data3[i].value); } return 0;}n = 6 0 0 0 0 0 0 0 0 0 4 0 0 0 0 0 0 0 0 0 0 0 0 3 10 0 4 0 10 6 0(1 3 4)(3 4 3)(3 5 10)(4 1 4)(4 3 10)(4 4 6) 3 0 0 0 9 0 0 0 7 0 0 0 0 4 0 0 0 0 10 0 0 0 0 0 0 0 0 0 0 4(0 0 3)(0 4 9)(1 2 7)(2 1 4)(3 0 10)(4 5 4) 3 0 0 0 9 0 0 0 7 4 0 0 0 4 0 0 0 0 10 0 0 0 3 10 0 4 0 10 6 4k = 0(0 0 3)(0 4 9)(1 2 7)(1 3 4)(2 1 4)(3 0 10)(3 4 3)(3 5 10)(4 1 4)(4 3 10)(4 4 6)(4 5 4)3.
编写一个算法，寻找m×n的矩阵中的每一行的最大值元素，并用三元组存储该最大值元素及其位置。
#include <stdio.h>#define M 3#define N 4struct { int row; int col; int value;} data[M];int main(){ int A[M][N] = {1,2,3,4,5,6,7,8,9,10,11,12}; int i,j; for(i=0;i<M;i++){ for(j=0;j<N;j++){ printf("%4d",A[i][j]); } printf("
"); } for(i=0;i<M;i++){ data[i].row = i; data[i].col = 0; data[i].value = A[i][0]; } for(i=0;i<M;i++){ for(j=0;j<N;j++){ if(A[i][j]>A[i][0]){ data[i].row = i; data[i].col = j; data[i].value = A[i][j]; } } } for(i=0;i<M;i++){ printf("(%d %d %d)
",data[i].row,data[i].col,data[i].value); } return 0;}4.
编写一个算法，实现4阶杨辉魔方（阴图）。
#include <stdio.h>#define N 4void swap(int *p,int *q);int main(){ int *p,*q; int A[N][N] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16}; int i,j; //初始矩阵 for(i=0;i<N;i++){ for(j=0;j<N;j++){ printf("%4d",A[i][j]); } printf("
"); } for(i=0;i<N/2;i++){ for(j=0;j<N;j++){ if((i==j)||(i+j==3)){ p = &A[i][j],q = &A[3-i][3-j]; swap(p,q); } } } //最终矩阵，用于验证 for(i=0;i<N;i++){ printf("
"); for(j=0;j<N;j++){ printf("%4d",A[i][j]); } } return 0;}void swap(int *p,int *q){ int t; t = *p; *p = *q; *q = t;}