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输入格式：
输入先后给出两个矩阵A和B。对于每个矩阵，首先在一行中给出其行数R和列数C，随后R行，每行给出C个整数，以1个空格分隔，且行首尾没有多余的空格。输入保证两个矩阵的R和C都是正数，并且所有整数的绝对值不超过100。
输出格式：
若输入的两个矩阵的规模是匹配的，则按照输入的格式输出乘积矩阵AB，否则输出Error: Ca != Rb，其中Ca是A的列数，Rb是B的行数。
输入样例1：
2 3 1 2 3 4 5 6 3 4 7 8 9 0 -1 -2 -3 -4 5 6 7 8
输出样例1：
2 4 20 22 24 16 53 58 63 28
输入样例2：
3 2 38 26 43 -5 0 17 3 2 -11 57 99 68 81 72
输出样例2：
Error: 2 != 3 #include<stdio.h> int main() { //定义所需的变量及数组1 int n1=0,m1=0; int
arr1[100][100]={0}; scanf("%d%d",&n1,&m1); for(int i=0;i<n1;i++) { for(int
j=0;j<m1;j++) { scanf("%d",&arr1[i][j]); } } //定义第二个变量与数组2 int n2=0,m2=0;
scanf("%d%d",&n2,&m2); int arr2[100][100]={0}; for(int i=0;i<n2;i++) { for(int
j=0;j<m2;j++) { scanf("%d",&arr2[i][j]); } } //判断是否能计算 int arr3[100][100]={0};
if(n2!=m1) { printf("Error: %d != %d",m1,n2); } else//此处进行计算 { for(int
i=0;i<n1;i++) { for(int j=0;j<m2;j++) { for(int k=0;k<m1;k++) {
arr3[i][j]+=arr1[i][k]*arr2[k][j];//找到规律 } } } //输出 printf("%d %d\n",n1,m2);
for(int i=0;i<n1;i++) { for(int j=0;j<m2;j++) { if(j!=m2-1) { printf("%d
",arr3[i][j]); } else { printf("%d",arr3[i][j]); } } printf("\n"); } } return
0; }
