UVA 1341 - Different Digits（数论)

UVA 1341 - Different Digits

题目链接

题意：给定一个正整数n，求一个kn使得kn上用的数字最少，如果相同，则输出值最小的

思路：

首先利用鸽笼原理证明最多需要2个数字去组成

设一个数字k，组成k,kk,kkk,kkkk... %n之后余数必然在0 - (n - 1)之间，所以必然能选出两个余数相等的数字相减为0，这个数字就是由0和k组成的。

因此只要考虑一个数字和两个数字的情况，去bfs，记忆化余数，因为余数重复必然形成周期了

代码：

#include <stdio.h>
#include <string.h>

const int INF = 0x3f3f3f3f;
const int N = 66666;
int n, num[2], vis
, ans, save
;
struct Queue {
int mod, pre, num, len;
Queue(){}
Queue(int mod, int pre, int num, int len) {
this->mod = mod;
this->pre = pre;
this->num = num;
this->len = len;
}
} q[N * 2];

void out(int now, int d) {
if (now == -1) return;
out(q[now].pre, d - 1);
save[d] = q[now].num;
}

int judge(int now, int d) {
if (now == -1) return 0;
int tmp = judge(q[now].pre, d - 1);
if (tmp != 0) return tmp;
if (save[d] == q[now].num) return 0;
else if (q[now].num < save[d]) return -1;
else return 1;
}

void bfs() {
int head = 0, tail = 0;
if (num[0] != 0) {
q[tail++] = Queue(num[0] % n, -1, num[0], 1);
vis[num[0] % n] = 1;
}
if (num[1] != -1 && num[1] != 0) {
q[tail++] = Queue(num[1] % n, -1, num[1], 1);
vis[num[1] % n] = 1;
}
while (head < tail) {
Queue now = q[head];
if (now.len > ans) return;
if (now.mod == 0) {
if (now.len <= ans) {
if (now.len != ans || judge(head, ans - 1) < 0) {
ans = now.len;
out(head, ans - 1);
}
}
}
Queue next;
for (int i = 0; i < 2; i++) {
if (num[i] == -1) continue;
next = Queue((now.mod * 10 + num[i]) % n, head, num[i], now.len + 1);
if (vis[next.mod]) continue;
vis[next.mod] = 1;
q[tail++] = next;
}
head++;
}
}

int main() {
while (~scanf("%d", &n) && n) {
ans = INF;
for (int i = 1; i < 10; i++) {
num[0] = i; num[1] = -1;
memset(vis, 0, sizeof(vis));
bfs();
}
if (ans == INF) {
for (int i = 0; i < 10; i++) {
for (int j = i + 1; j < 10; j++) {
num[0] = i; num[1] = j;
memset(vis, 0, sizeof(vis));
bfs();
}
}
}
for (int i = 0; i < ans; i++)
printf("%d", save[i]);
printf("\n");
}
return 0;
}