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
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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 ( 10X1, 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 ));
    }
}