一. 自定义一串数字求 参数个数,最大值,最大值()---------方法一: def max(*a): m=a[0] p=a[0] n=0 for x in a: if x>m: m=x n+=1 for x in a: if x<p: p=x return n,m,pif __name__ == '__main__': list=max(3,4,5) print("参数个数{},最大值{},最小值{}".format(list[0],list[1],list[2]))
Choose the best route Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7062 Accepted Submission(s): 2301 Problem Description One day , Kiki wants to visit one of her friends. As she is liable
在数组a中,a[i]+a[j]=a[k],求a[k]的最大值,a[k]max. 思路:将a中的数组两两相加,组成一个新的数组.并将新的数组和a数组进行sort排序.然后将a数组从大到小与新数组比较,如果当比较到a中第二个数组时,仍无满足条件,则返回最大值不存在. 情况一:不考虑i和j相等的情况.此时新数组长度为a.length*(a.length-1)/2; import java.util.Arrays; public class max { public static void main(S
# coding=utf-8 #共轭梯度算法求最小值 import numpy as np from scipy import optimize def f(x, *args): u, v = x a, b, c, d, e, f,g,h = args return a*u**g+ b*u*v + c*v**h + d*u + e*v + f def gradf(x, *args): u, v = x a, b, c, d, e, f,g,h = args gu = g*a*u + b*v +
Testing the CATCHER Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 16131 Accepted: 5924 Description A military contractor for the Department of Defense has just completed a series of preliminary tests for a new defensive missile calle
下面举例说明如何运用GA工具箱求解多约束非线性规划问题: function f =fitness(x) f=exp(x(1))*(4*x(1)^2+2*x(2)^2+4*x(1)*x(2)+2*x(2)+1); fitness.m function [c,ceq]=constraint(x) c(1)=1.5+x(1)*x(2)-x(1)-x(2); c(2)=-x(1)*x(2)-10; ceq=[]; %一定要有 constraint.m 注意integer variable indices
求模和求余的总体计算步骤如下: 1.求整数商 c = a/b 2.计算模或者余数 r = a - c*b 求模和求余的第一步不同,求余在取c的值时向0方向舍入;取模在计算c的值时向无穷小方向舍入. C语言实现 //取余 int rem(int a, int b) { int c = a * 1.0 / b; return (a - c * b); } //求模 int mod(int a, int b) { int c = floor(a * 1.0 / b); //#include <mat