紫书 例题7-1 UVA 725

【题意】 给你一个数字n，用0~9,10个数字组成两个五位数，使得他们的商为n，按顺序输出所有结果。

【解题方法】暴力。直接枚举第二个数字，范围（1000,100000），然后判断即可。

【AC代码】

//
//Created by just_sort 2016/1/3
//Copyright (c) 2016 just_sort.All Rights Reserved
//

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <cstdio>
#include <time.h>
#include <cstdlib>
#include <cstring>
#include <sstream> //isstringstream
#include <iostream>
#include <algorithm>
using namespace std;
using namespace __gnu_pbds;
typedef long long LL;
typedef pair<int, LL> pp;
#define REP1(i, a, b) for(int i = a; i < b; i++)
#define REP2(i, a, b) for(int i = a; i <= b; i++)
#define REP3(i, a, b) for(int i = a; i >= b; i--)
#define CLR(a, b)     memset(a, b, sizeof(a))
#define MP(x, y)      make_pair(x,y)
const int maxn = 100;
const int maxm = 2e5;
const int maxs = 10;
const int INF  = 1e9;
const int UNF  = -1e9;
const int mod  = 1e9+7;
int gcd(int x, int y) {return y == 0 ? x : gcd(y, x % y);}
//typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>order_set;
//head

int main()
{
    int n, ks = 0;
    while(scanf("%d", &n) != EOF && n){
        if(ks++) printf("\n");
        bool ok = 0;
        //int st = clock();
        REP1(i, 0, 10){
            REP1(j, 0, 10){
                REP1(k, 0, 10){
                    REP1(l, 0, 10){
                        REP1(m, 0, 10){
                            if(i != j && i != k && i != l && i != m && j != k && j != l && j != m && k != l && k != m
                               && l != m){
                                int t1 = i * 10000 + j * 1000 + k * 100 + l * 10 + m;
                                if(t1 % n == 0){
                                    int t2 = t1 / n;
                                    int a[10], cnt = 0, b[5];
                                    while(t2){
                                        a[cnt++] = t2 % 10;
                                        t2 /= 10;
                                    }
                                    while(cnt < 5) a[cnt++] = 0;
                                    b[0] = a[4];
                                    b[1] = a[3];
                                    b[2] = a[2];
                                    b[3] = a[1];
                                    b[4] = a[0];
                                    a[cnt++] = i, a[cnt++] = j, a[cnt++] = k, a[cnt++] = l, a[cnt++] = m;
                                    sort(a, a + 10);
                                    bool flag = true;
                                    REP1(ii , 0, 9){
                                        if(a[ii] == a[ii + 1]){
                                            flag = false;
                                        }
                                    }
                                    if(flag){
                                        ok = 1;
                                        printf("%d%d%d%d%d", i, j, k, l, m);
                                        printf(" / ");
                                        printf("%d%d%d%d%d", b[0], b[1], b[2], b[3], b[4]);
                                        printf(" = %d\n", n);
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
        if(ok == 0){
            printf("There are no solutions for %d.\n", n);
        }
        //int en = clock();
        //cout << en - st << endl;
    }
    return 0;
}