uva 1341 - Different Digits(数论+bfs)

题目链接：uva 1341 - Different Digits

题目大意：给定一个数字n，要求求一个数字m，m可以整除n，并且尽量组成的数字种类(0~9)尽量少，在种类相同的情况下数值尽量小。

解题思路:可以证明两种数字肯定可以组成m，假设有数字k，一种数字可以有k，kk，kkk，kkkk，…整除n剩的数一定在0~n-1之间，所以肯定存在两个由k数字组成的数字同模，那么这两个数相减所得到的数即使kkk00000，两种数字。于是肯定了范围，枚举数字，然后用bfs获取答案，维护最小值即可。

#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>

using namespace std;

const int maxn = 65536;
int n, vis[maxn+10];

struct state {
int key, len, vec[maxn];
state () {
key = 0;
}

bool operator < (const state& a) {
if (a.key == 0)
return true;
if (key == 0)
return false;

if (key != a.key)
return key < a.key;

if (len != a.len)
return len < a.len;

int l = len;
for (int i = 0; i < l; i++)
if (vec[i] != a.vec[i])
return vec[i] < a.vec[i];
return false;
}
};

struct node {
int mod, len, num;
void set (int mod, int num, int len) {
this->mod = mod;
this->num = num;
this->len = len;
}
}que[maxn*2];

void put (int d, state& w, int x) {
if (d < 0 || x < 0)
return;
w.vec[d] = que[x].num;
put(d-1, w, vis[que[x].mod]);
}

state judge (int a, int b) {
int head = 1, rear = 1;
memset(vis, -1, sizeof(vis));

state w;
node u, v;
w.key = (a==b?1:2);

if (a) {
que[rear++].set(a%n, a, 1);
vis[a%n] = 0;
}

if (b) {
que[rear++].set(b%n, b, 1);
vis[b%n] = 0;
}

while (head < rear) {
u = que[head++];

if (u.mod == 0) {
w.len = u.len;
put(u.len-1, w, head-1);
return w;
}

for (int i = 0; i < w.key; i++) {
int t = (i?b:a);
v.set((u.mod*10+t)%n, t, u.len+1);

if (vis[v.mod] != -1)
continue;

vis[v.mod] = head-1;;
que[rear++] = v;
}
}
w.key = 0;
return w;
}

int main () {
while (scanf("%d", &n) == 1 && n) {
state ans;

for (int i = 1; i <= 9; i++) {
state c = judge(i, i);
if (c < ans)
ans = c;
}

if (ans.key == 0) {
for (int i = 0; i <= 9; i++) {
for (int j = i+1; j <= 9; j++) {
state c = judge(i, j);
if (c < ans)
ans = c;
}
}
}
for (int i = 0; i < ans.len; i++)
printf("%d", ans.vec[i]);
printf("\n");
}
return 0;
}