zoukankan      html  css  js  c++  java
  • HNUSTOJ-1565 Vampire Numbers(暴力打表)

    1565: Vampire Numbers

    时间限制: 3 Sec  内存限制: 128 MB
    提交: 20  解决: 9
    [提交][状态][讨论版]

    题目描述

    The number 1827 is an interesting number, because 1827=21*87, and all of the same digits appear on both sides of the `='. The number136948 has the same property: 136948=146*938.

    Such numbers are called Vampire Numbers. More precisely, a number v is a Vampire Number if it has a pair of factors, a and b, wherea*b = v, and together, a and b have exactly the same digits, in exactly the same quantities, as v. None of the numbers v, a or b can have leading zeros. The mathematical definition says that v should have an even number of digits and that a and b should have the same number of digits, but for the purposes of this problem, we'll relax that requirement, and allow a and b to have differing numbers of digits, and v to have any number of digits. Here are some more examples:

    126 = 6 * 21
    10251 = 51 * 201
    702189 = 9 * 78021
    29632 = 32 * 926
    

    Given a number X, find the smallest Vampire Number which is greater than or equal to X.

    输入

    There will be several test cases in the input. Each test case will consist of a single line containing a single integer X ( 10$ le$X$ le$1, 000, 000). The input will end with a line with a single `0'.

    输出

    For each test case, output a single integer on its own line, which is the smallest Vampire Number which is greater than or equal to X. Output no extra spaces, and do not separate answers with blank lines.

    样例输入

    10
    126
    127
    5000
    0

    样例输出

    126
    126
    153
    6880
    #include<algorithm>
    #include<iostream>
    #include<cstring>
    #include<cstdio>
    #include<set>
     
    using namespace std;
    int vis[10];
    set<int> S;
    void Init_set(){
        for(int i = 1; i <= 1010; i++)
            for(int j = i; j <= 1000500/i; j++){
                int m = i * j;
                int tx = i, ty = j, tm = m;
                while(tm) {vis[ tm%10 ]++; tm /= 10;}
                while(tx) {vis[ tx%10 ]--; tx /= 10;}
                while(ty) {vis[ ty%10 ]--; ty /= 10;}
                bool is = true;
                for(int i = 0; i < 10; i++)
                    if(vis[i]) is = false, vis[i] = 0;
                if(is) S.insert( m );
            }
    }
    int main(){
        int x;
        Init_set();
        while(scanf("%d", &x) == 1 && x){
            printf("%d
    ", *S.lower_bound( x ));
        }
    }
     
  • 相关阅读:
    挺好的 "简历修改"
    请问谁会破解QQ空间相册的密码呀教我下谢谢!!。
    SEO工具,站长必备
    解决vs2005自带水晶报表次数的限制的次数
    虚拟主机如何实现泛域名解析?
    读出数据库里面的数据,来 绑定列表框 DropDownList1
    取出数据里面相同的记录
    如何让文本框输入字母自动换行???? 很有意思的哦 
    net事件丢失解决方法
    暑假集训每日一题0727 (网络流)
  • 原文地址:https://www.cnblogs.com/Pretty9/p/7406651.html
Copyright © 2011-2022 走看看