Ultraviolet A 12009-avaricious maryanna (violent)

Connection: Ultraviolet A 12009-avaricious maryanna

Given n, x and X are n digits, and the last n digits of x * X are still X.

Solution: When you create a table, you will find that there are actually regular rules. Except when n = 1, there are 0 and 1 more. The other values are n-1 digits, and then a new number of digits is added, the first digit is 5, 6.

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 500;int a[maxn+5], b[maxn+5];int mul (int* p, int n) {    int q[maxn+5];    memset(q, 0, sizeof(q));    for (int i = 1; i < n; i++) {        int t = 0, flag = true;        for (int j = 1; j < n; j++) {            if (i + j - 1 > n) {                flag = false;                break;            }            t += p[i] * p[j] + q[i+j-1];            q[i+j-1] = t % 10;            t /= 10;        }        if (flag) {            int mv = i+n-1;            while (t) {                q[mv++] = t % 10;                t /= 10;            }        }    }    /*       for (int i = 1; q[i]; i++)       printf("%d", q[i]);       printf("\n");       */    return q[n];}void init () {    a[1] = 5;    b[1] = 6;    mul(a, 2);    for (int i = 2; i <= maxn; i++) {        int ra = 0, rb = 0;        int p = mul(a, i);        int q = mul(b, i);        for (int k = 0; k <= 9; k++) {            if ((2 * k * a[1] + p) % 10 == k)                ra = k;            if ((2 * k * b[1] + q) % 10 == k)                rb = k;        }        a[i] = ra;        b[i] = rb;    }}void put (int* num, int n) {    printf(" ");    for (int i = n; i; i--)        printf("%d", num[i]);}int main () {    init();    int cas, n;    scanf("%d", &cas);    for (int k = 1; k <= cas; k++) {        scanf("%d", &n);        printf("Case #%d:", k);        if (n == 1)            printf(" 0 1");        if (a[n] && b[n]) {            if (a[n] > b[n]) {                put(b, n);                put(a, n);            } else {                put(a, n);                put(b, n);            }        } else if (a[n]) {            put(a, n);        } else if (b[n])            put(b, n);        printf("\n");    }    return 0;}