紫书入门级别的暴力，可我还是写了好长时间 = =

【题目链接】uva 725

【题目类型】化简暴力

&题解：
首先要看懂题意，他的意思也就是0~9都只出现一遍，在这2个5位数中。
接着，你要知道：枚举一个5位数就够了，不用2个5位数都枚举，因为你可以通过n知道第2个5位数。
最后set维护出现的次数，ok判断是否可行，pri输出。

【时间复杂度】O(1e5)

&代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
#define cle(a,val) memset(a,(val),sizeof(a))
#define SI(N) scanf("%d",&(N))
#define SII(N,M) scanf("%d %d",&(N),&(M))
#define SIII(N,M,K) scanf("%d %d %d",&(N),&(M),&(K))
#define rep(i,b) for(int i=0;i<(b);i++)
#define rez(i,a,b) for(int i=(a);i<=(b);i++)
#define red(i,a,b) for(int i=(a);i>=(b);i--)
const ll LINF = 0x3f3f3f3f3f3f3f3f;
#define PU(x) puts(#x);
int n;
set<int> sei, se2;
vector<pair<int, int> > vep;
bool ok(int d) {se2 = sei;int t = d / n;if (t * n != d) {return false;}int u = 0;while (t) {u++;sei.insert(t % 10);t /= 10;}if (u > 5) {return false;}if (sei.size() == 9 && !sei.count(0) && u == 4) {return true;}if (sei.size() == 10) {return true;}return false;
}
void pri() {if (vep.empty()) {printf("There are no solutions for %d.\n", n);return;}int t = vep.size();rep(i, t)printf("%05d / %05d = %d\n", vep[i].first, vep[i].second, n);
}
void Solve() {int uu = 0;while (~SI(n), n) {if (uu) PU()uu = 1;sei.clear() ;vep.clear();rez(i1, 0, 9) {sei.insert(i1);rez(i2, 0, 9) {if (sei.count(i2)) continue;sei.insert(i2);rez(i3, 0, 9) {if (sei.count(i3)) continue;sei.insert(i3);rez(i4, 0, 9) {if (sei.count(i4)) continue;sei.insert(i4);rez(i5, 0, 9) {if (sei.count(i5)) continue;sei.insert(i5);int d = i1 * 1e4 + i2 * 1e3 + i3 * 1e2 + i4 * 1e1 + i5;if (ok(d)) {pair<int, int> p;p.first = d;p.second = d / n;vep.push_back(p);}sei = se2;sei.erase(i5);}sei.erase(i4);}sei.erase(i3);}sei.erase(i2);}sei.erase(i1);}pri();}
}
int main() {Solve();return 0;
}

上面是按位搜索的暴力，代码较长，下面是直接枚举的暴力，枚举枚举范围1234~98765，之后再判断，这样写代码就较短了。我枚举的是第一个5位数，当然，也可以枚举第二个，你自己可以试下。

&代码2：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
#define cle(a,val) memset(a,(val),sizeof(a))
#define SI(N) scanf("%d",&(N))
#define SII(N,M) scanf("%d %d",&(N),&(M))
#define SIII(N,M,K) scanf("%d %d %d",&(N),&(M),&(K))
#define rep(i,b) for(ll i=0;i<(b);i++)
#define rez(i,a,b) for(ll i=(a);i<=(b);i++)
#define red(i,a,b) for(ll i=(a);i>=(b);i--)
const ll LINF = 0x3f3f3f3f3f3f3f3f;
#define PU(x) puts(#x);
#define PI(A) cout<<(A)<<endl;
#define DG(x) cout<<#x<<"="<<(x)<<endl;
#define DGG(x,y) cout<<#x<<"="<<(x)<<" "<<#y<<"="<<(y)<<endl;
#define DGGG(x,y,z) cout<<#x<<"="<<(x)<<" "<<#y<<"="<<(y)<<" "<<#z<<"="<<(z)<<endl;
#define PIar(a,n) rep(i,n)cout<<a[i]<<" ";cout<<endl;
#define PIarr(a,n,m) rep(aa,n){rep(bb, m)cout<<a[aa][bb]<<" ";cout<<endl;}
const double EPS = 1e-9 ;
/*     C o d i n g  S p a c e     */
const int MAXN = 1000 + 5 ;
int n;
bool used[10];
bool ok(int x, int y) {if (y*n!=x) return false;cle(used, 0);//这是1e4 不是1e5if (x < 10000) used[0] = 1;if (y < 10000) used[0] = 1;int u1 = 0, u2 = 0;while (x) {used[x % 10] = 1;x /= 10;u1++;}while (y) {used[y % 10] = 1;y /= 10;u2++;}int c = 0;if (u1 < 6 && u2 < 6)rep(i, 10) if (used[i]) c++;return (c == 10);
}
void Solve() {int kg = 0;while (~SI(n), n) {if (kg)puts("");kg = -1;for (int i = 1234; i <= 98765; i++)if (ok(i, i / n)) {printf("%05d / %05d = %d\n", i, i / n, n);kg = 1;}if (kg == -1) printf("There are no solutions for %d.\n", n);}
}
int main() {Solve();return 0;
}