主要在于判断是否能被整除,思路是用取余运算符%,取余结果为0就表示能被整除. 代码如下: public class NumDemo { public static void main(String args[]){ int n; System.out.println("在1~1000可被3与5整除的为"); for(n=1;n<=1000;n++){ if(n%3==0&&n%5==0) { System.out.println("1~1000之间能够同…
B-number Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3257 Accepted Submission(s): 1819 Problem Description A wqb-number, or B-number for short, is a non-negative integer whose decimal for…
#include<bits/stdc++.h> #define ll long long using namespace std; int m; int n,p[20]; int sum,ans; int main() { scanf("%d%d",&n,&m); for(int i=1;i<=m;i++) scanf("%d",&p[i]); for(int i=1;i<1<<m;i++) { int c…
public class Test21 { public static void main(String[] args) { // TODO Auto-generated method stub int sum = 0; int[] array = new int[sum]; int index=0; for (int i = 1; i < 100; i++) { if (i % 3…
分别用while,do-while,for语句实现 方法一:while #include<stdio.h> int main() { int m; m = ; ) { == ) printf("%5d", m); m = m + ; } } 方法二:do-whlie #include<stdio.h> int main() { int m; m = ; do { == ) printf("%5d", m); m = m + ; } ); }…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5373 The shortest problem Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 995 Accepted Submission(s): 498 Problem Description In this problem, w…
题目链接: http://acm.hust.edu.cn/vjudge/problem/47664 Eleven Time Limit: 5000MS 问题描述 In this problem, we refer to the digits of a positive integer as the sequence of digits required to write it in base 10 without leading zeros. For instance, the digits o…
