c语言编程题

#include "stdafx.h"
#include <stdio.h>
#include <malloc.h>
int main(int argc, char* argv[])
{
    
	int i, number, *p;
	scanf("%d", &number); // 输入向量长度
	p = (int*)malloc(sizeof(int)*number*2); // 申请长度为number个int的连续存储空间
	printf("请输入长度为%d的两个向量:\n",number);
	for (i = 0; i < number*2; i++)
	{
    
		scanf("%d", &p[i]); // 获取输入流中的数据（以int进行解析）
	}
	int result = 0;
	// 计算向量内积
	for (i = 0; i < number; i++)
	{
    
		result+= (p[i] * p[i+number]);
	}
	printf("result is %d\n", result);
	return 0;
}

#include "stdafx.h"
#include <stdio.h>
#include <malloc.h>
struct RowMin
{
    
	int value;
	int row;
	int col;
};
int main(int argc, char* argv[])
{
    
	int i, number, *p;
	scanf("%d", &number);
	p = (int*)malloc(sizeof(int)*number*number); // 申请内存空间n*n长度
	printf("请输入长度为%d*%d的数组:\n",number,number);
	for (i = 0; i < number*number; i++)
	{
    
		scanf("%d", &p[i]);
	}
	// 找出每一行的最小值
	RowMin *pRowMin = (RowMin*)malloc(sizeof(RowMin)*number); // 存储每一行的最小值
	for (i = 0; i < number; i++)
	{
    
		pRowMin[i].value = p[number*i];
		pRowMin[i].row = i;
		pRowMin[i].col = 0;
		for (int j = 1; j < number; j++)
		{
    
			if (p[number*i+j] < pRowMin[i].value)
			{
    
				pRowMin[i].value = p[number*i+j];
				pRowMin[i].col = j;
			}
			
		}
	}
	// 找出每一行的最大值
	int max = 0;
	for (i = 1; i < number; i++)
	{
    
		if (pRowMin[max].value < pRowMin[i].value)
		{
    
			max = i;
		}
	}

	printf("result is %d,%d,%d\n", pRowMin[max].value, pRowMin[max].row, pRowMin[max].col);
	return 0;
}