虽然是暴力求解,但是也要观察条件,尽量提高效率.如本题,原本要枚举10个数,但是分析可知通过枚举fghij就可以了. #include<stdio.h> #include<string.h> //判断a,b中的所有数字都不等 int judge(int a,int b){ ) ; ]; ; int i; memset(ans,,sizeof(ans)); ) ans[]=;//因为如果是四位数,要考虑第一个0 while(a){ ans[a%]=; a/=; } while(b){…

#include<cstdio> #include<cstring> #include<algorithm> using namespace std; ; int vis[maxn]; int check(int n, int x) { memset(vis, , sizeof(vis)); int a, t; t = n / x; ) vis[] = ; //这步极为关键,刚开始没有这步,测试样例都过不了 //找了半天才发现.其实,如果出现前导0,一定是在 //分母,…

题目链接>>>>>> 题目大意:给你一个数n(2 <= n <= 79),将0-9这十个数字分成两组组成两个5位数a, b(可以包含前导0,如02345也算),使得a / b = n:列出所有的可能答案. #include <iostream> #include <cstdio> #include <cstring> #include <string> #include <algorithm> #i…

uva 725 Division Write a program that finds and displays all pairs of 5-digit numbers that between them use the digits 0 through 9 once each, such that the first number divided by the second is equal to an integer N, where 2<=N <=79. That is, abcde…

Description Write a program that finds and displays all pairs of 5-digit numbers that between them use the digits 0 through 9 once each, such that the first number divided by the second is equal to an integer N, where . That is, abcde / fghij =N wher…

如果把数字0到9分配成2个整数(各五位数),现在请你写一支程序找出所有的配对使得第一个数可以整除第二个数,而且商为N(2<=N<=79),也就是:abcde / fghijk = N这里每个英文字母代表不同的数字,第一个数字可以为0.Input输入包含许多笔待测数据,每列代表一笔待测数据,每笔待测数据包含一个正整数N,N为0时代表输入结束.Output对每笔待测数据由小到大输出每一对符合条件的数.如果找不到符合条件的数对,则输出There are no solutions for N..每笔测…

除法(Division,uva725) 输入整数n,按从小到大的顺序输出所有形如abcde/fghij=n的表达式,其中a~j恰好为数字0~9的一个排列(可以有前导0),2<=n<=79. 输入:输入若干组数据,以文件结束符结束. 输出:For each test case you must print the message: Case #M: The maximum product is P., where M is the number of the test case, startin…

题目链接:https://vjudge.net/problem/UVA-725 Write a program that finds and displays all pairs of 5-digit numbers that between them use the digits 0 through 9 once each, such that the first number divided by the second is equal to an integer N, where 2 ≤…

//Uva725 #include <iostream> #include <cstring> #include <cstdlib> #include <cstdio> using namespace std; void evalu(int n) { + ; ]; //将数字用字符保存 ]; //判断每个数,是否重复 char buf[maxn]; //将出现的字符全部存到里面 ; i <= ; i++) { memset(flag, , sizeof…