写了一段相关的代码,好久不写了#include <iostream>#include <exception>#include <stack>using namespace std;int func(int a, int b) {
https://www.u72.net/daima/4741.html - 2024-09-05 15:55:51 - 代码库获得两个随机数(100以内),并放入数组中public int[] getTwoRandom(){ int[] t = new int[2]; Random rand = new Random(); for(int i=0;i
https://www.u72.net/daima/85wn.html - 2024-07-26 18:37:52 - 代码库【算法】欧几里德算法#include<cstdio>int gcd(int a,int b){return b==0?a:gcd(b,a%b);}int main(){ int a,b; scanf("%d%d",&a,&b);
https://www.u72.net/daima/nkf8k.html - 2024-09-26 04:21:02 - 代码库#include<stdio.h>int main(){ int a,b,r; scanf("%d%d",&a,&b); r=a%b; while(r!=0) { a=b; b=r; r=
https://www.u72.net/daima/nhfb8.html - 2024-08-02 15:29:43 - 代码库题目背景全场基本暴力题目描述输入输出格式输入格式: 如图 输出格式: 如图 输入输出样例输入样例#1:如图输出样例#1:如图说明如图 这题用到了容斥原理和
https://www.u72.net/daima/numbh.html - 2024-10-27 08:17:39 - 代码库Time Limit: 10 Sec Memory Limit: 256 MBSubmit: 172 Solved: 101Description给定一个长度为 N 的正整数序列Ai对于其任意一个连续的子序列{Al,Al
https://www.u72.net/daima/ndamf.html - 2024-09-28 18:55:38 - 代码库//algorithm.henum SWAP_TYPE{MEMORY, COMPLEX};struct SIntArray{ int *pData; int num; SIntArray():pData(NULL),num(0){} void Clear
https://www.u72.net/daima/sswn.html - 2024-07-13 01:49:08 - 代码库之前一直只知道欧几里得辗转相除法,今天学习了一下另外一种、在处理大数时更优秀的算法——Stein特此记载 1.欧几里得(Euclid)算法又称辗转相除法,依据
https://www.u72.net/daima/x19z.html - 2024-08-27 14:11:02 - 代码库求最小<em>公约数</em>,最容易想到的是欧几里得算法,这个算法也是比较容易理解的,效率也是很不错的。也叫做辗转相除法。
https://www.u72.net/daima/nzu7u.html - 2024-08-01 18:33:27 - 代码库/*Time Limit : 1000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)Total Submission(s) : 148 Accepted Submission(s) : 19
https://www.u72.net/daima/uske.html - 2024-07-14 02:37:29 - 代码库package javaapplication24;class NegativeIntegerException extends Exception{String message;public NegativeIntegerException(){message="方法的
https://www.u72.net/daima/xzc8.html - 2024-07-16 19:07:57 - 代码库#include<stdio.h>int main(){ int a,b;//两个数 int x,y;//暂时保存a,b以便后面求最小公倍数p=(a*b)/a int p;//最小公倍数 int temp; while(s
https://www.u72.net/daima/nhawk.html - 2024-08-02 08:56:18 - 代码库1 import java.util.Scanner; 2 3 public class ZuiDaGongYueShuClass { 4 5 public static void main(String[] args) { 6 Scan
https://www.u72.net/daima/nwhwb.html - 2024-11-04 06:23:39 - 代码库Description输入二个正整数x0,y0(2<=x0<100000,2<=y0<=1000000),求出满足下列条件的P,Q的个数条件: 1.P,Q是正整数2.要求P,Q以x0为<em>最大</em>
https://www.u72.net/daima/ncc9k.html - 2024-10-10 07:17:01 - 代码库1.求<em>最大公约数</em>辗转相除求值例如:12,8求<em>最大公约数</em>12%8=4[不为零]8%4=0[为零,则4为<em>最大公约数</em>]【原理就是12,8的<em>最大公约数</em>和
https://www.u72.net/daima/nub7z.html - 2024-10-22 10:13:02 - 代码库a=int(raw_input(‘a‘))b=int(raw_input(‘b‘))su=[]if a>b: smaller=b else: smaller=a for i in range(1,smaller+1): if
https://www.u72.net/daima/nc5w8.html - 2024-10-11 19:00:39 - 代码库#include<stdio.h>int gcd(int a,int b){ if(b!=0) gcd(b,a%b); else return a;}int lcm(int a,int b){ return a*b/gcd(a,b); //a/gcd(a,b
https://www.u72.net/daima/nfkre.html - 2024-10-06 07:10:39 - 代码库[摘要]一个朋友在展BOM的时候有这样的需求,两列字段(数值):A ,B A=用量,B=底数,组成用量=用量/底数。A/B,若能被整除,显示整除的结果,若不能整除显示分
https://www.u72.net/daima/dkw6.html - 2024-07-07 17:41:39 - 代码库链接: http://soj.me/1732ConstraintsTime Limit: 1 secs, Memory Limit: 32 MBDescription:Alice is a beautiful and clever girl. Bob would li
https://www.u72.net/daima/zacr.html - 2024-07-04 10:25:43 - 代码库# include <stdio.h>int main(void){ int x, y,temp; int r; printf("请输入两个正整数:\n"); scanf("%d %d", &num1, &num2); r = num1 % num2; temp
https://www.u72.net/daima/s93s.html - 2024-07-13 15:02:31 - 代码库