Sequence (opentrains)

这是一道找规律的题目
题意是生成一个序列，后一个数是前一个数的两倍，再通过每一位数进行排序生成的新的一个数
一开始想如果出现了5，那么极有可能生成0，使得新的数的位数减少，所以可能存在一个循环节，然后我通过打表出前1000个数，很容易就找到规律

代码：

#include <bits/stdc++.h>using namespace std;int a[10], b[1000][10], t, n, m;const int mo = 2147483600;int main() {    //freopen("c.out","w",stdout);    freopen("input.txt","r",stdin);    freopen("output.txt","w",stdout);    scanf("%d", &n);    m = n;    a[1] = 1;    memset(b, 0, sizeof(b));    b[1][1] = 1;    for (int x = 1; x <= 1000; x++) {        /*if (x) {            //for (int i = 0; i < 9; i++) printf("%d,", a[i]);            //printf("%d",a[9]);            for (int i = 1; i <= 9; i++)                for (int j = 1; j <= a[i]; j++) printf("%d", i);            printf("\n");        }*/        t = 0;        for (int i = 9; i >= 1; i--) if (a[i]) {            if (i >= 5) {                b[x+1][(i*2+1)%10] += a[i]-1;                if (t) b[x+1][(i*2+1)%10]++;                else b[x+1][(i*2)%10]++;            } else {                b[x+1][(i*2)%10] += a[i]-1;                if (t) b[x+1][(i*2+1)%10]++;                else b[x+1][(i*2)%10]++;            }            t = (i*2)/10;        }        if (t) b[x+1][1]++;        for (int i = 0; i <= 9; i++) a[i] = b[x+1][i];    }    n = m;    if (n > 1000) n = (n-37)%6+37;    for (int i = 1; i <= 9; i++)        for (int j = 1; j <= b[n][i]; j++) printf("%d", i);    printf("\n");}