求平均成绩 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 58415 Accepted Submission(s): 13985 Problem Description 假设一个班有n(n<=50)个学生,每人考m(m<=5)门课,求每个学生的平均成绩和每门课的平均成绩,并输出各科成绩均大于等于平均成绩的学生数量. Inpu
求平均成绩 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 74055 Accepted Submission(s): 17809 Problem Description 假设一个班有n(n<=50)个学生,每人考m(m<=5)门课,求每个学生的平均成绩和每门课的平均成绩,并输出各科成绩均大于等于平均成绩的学生数量. Inpu
求平均成绩 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 61842 Accepted Submission(s): 14812 Problem Description 假设一个班有n(n<=50)个学生,每人考m(m<=5)门课,求每个学生的平均成绩和每门课的平均成绩,并输出各科成绩均大于等于平均成绩的学生数量. Inp
Home Web Board ProblemSet Standing Status Statistics Problem F: 求平均年龄 Problem F: 求平均年龄 Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 720 Solved: 394[Submit][Status][Web Board] Description 定义一个Persons类,用于保存若干个人的姓名(string类型)和年龄(int类型),定义其方法 void ad
MATLAB中求矩阵非零元的坐标: 方法1: index=find(a); [i,j]=ind2sub(size(a),index); disp([i,j]) 方法2: [i,j]=find(a>0|a<0) %列出所有非零元的坐标 [i,j]=find(a==k) %找出等于k值的矩阵元素的坐标 所用函数简介: IND2SUB Multiple subscripts from linear index. IND2SUB is used to determine the equivalent
quake3中求1/sqrt(x)的算法源代码如下(未作任何修改): float Q_rsqrt( float number ) { long i; float x2, y; const float threehalfs = 1.5F; x2 = number * 0.5F; y = number; i = * ( long * ) &y; // evil floating point bit level hacking i = ); // what the fuck? y = * ( floa