Seeyouer的博客

暴力枚举： i从10000开始找（因为N最小为2，所以i一定大于i/N，一定是五位数），到98765，如果是N的倍数就开始判断i和i/N是否为不重复数，是->输出，不是->继续找；注意： ① 0的位置及个数，如果i/N是四位数，那么0要添加在分母前面； ②输出格式格式格式，我就卡这了，输入0之后不输出空行；代码：#include<iostream>#include<stdio.h>#inc

看清楚题意，连续的子序列，不能从小到大排序再乘。#include<iostream>#include<stdio.h>#include<string.h>#include<string>#include<math.h>#include<algorithm>using namespace std;int main(){ long long max,max1;//注意范围

Sample Input -190 -191 -192 -193 -194 195 196 197 198 199 200 0 Sample Output -190 = -1 x 2 x 5 x 19 -191 = -1 x 191 -192 = -1 x 2 x 2 x 2 x 2 x 2 x 2 x

Floating-point numbers are represented differently in computers than integers. That is why a 32-bit floating-point number can represent values in the magnitude of 1038 while a 32-bit integer can

题目描述Keeping an eye on long term career possibilities beyond the farm, Bessie the cow has started learning algorithms from various on-line coding websites. Her favorite algorithm thus far is “bubble...