//求出4×4矩阵中最大和最小元素值及其所在行下标和列下标,求出两条主对角线元素之和 #include <stdio.h> int main() { int sum=0; int max,min; int max1,max2;//记录最大值的坐标 int min1,min2;//记录最小值的坐标 int i,j; int a[4][4]; //为数组赋值 for(i=0;i<4;i++) { for(j=0;j<4;j++) { scanf("%d",&…
找出所有点: 根据斜率按照一个方向递增,求出对应的另一个方向的整数值. Point pStart = new Point(0, 2); Point pEnd = new Point(8, 2); //定义线上的点 List<Point> linePoint = new List<Point>(); //定义x坐标的大小 Point pointMaxX = new Point(); Point pointMinX = new Point(); //给x坐标大的点和小的点赋值 if (…
find the longest of the shortest Time Limit: 1000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2524 Accepted Submission(s): 888 Problem Description Marica is very angry with Mirko because he found a new gi…