Uva10273-eat or not to eat? (Violent)

Uva10273-eat or not to eat? (Violent)

Question Link

In order to improve income, a farm owner decides to kill the ox with the least yield every day. But if one day has more than one ox with the least yield, so this day will not be killed. Then I will give you a nheaded ox, give you the yield cycle and the production in the cycle of each ox, and ask how many days later you can determine the status of the remaining ox.

Solution: Find the minimum public multiple of the cycle of the native ox. If no ox is killed in this cycle, the subsequent situation will be cyclical, and no more cattle will be killed, in this case, the situation is determined. Under such conditions, the force will not time out.

Code:

#include <cstdio>#include <cstring>using namespace std;const int N = 1005;const int M = 15;const int maxn = 255;int Milk[N][M];int cnt[N];int vis[N];int survice[N];int n;int gcd (int a, int b) {    return b == 0 ? a : gcd (b, a % b);}int f() {    int T = 1;    int tmp;    for (int i = 0; i < n; i++) {        if (vis[i])            continue;        if (T > cnt[i])            tmp = gcd(T, cnt[i]);        else            tmp = gcd(cnt[i], T);        T = T * cnt[i] / tmp;    }    return T;}int solve () {    int T = f();    memset (vis, 0, sizeof (vis));    //    memset (survice, -1, sizeof (survice));    survice[n] = 0;    int num = n;    for (int p = 0;; p++) {        bool flag = 0;        for (int i = 1; i <= T; i++) {            int minnum = maxn;            int tmp = 0;            for (int j = 0; j < n; j++) {                if (vis[j])                     continue;                int d = (i % cnt[j]) == 0 ? cnt[j] : (i % cnt[j]);                if (minnum > Milk[j][d]) {                    minnum = Milk[j][d];                    tmp = j;                } else if (minnum == Milk[j][d])                    tmp = -1;                }            if (tmp != -1) {                vis[tmp] = 1;                num--;                survice[num] = p * T + i;;                flag = 1;            }        }        if (!flag)            return num;    }}int main () {    int cas, d;    scanf ("%d", &cas);    while (cas--) {        scanf ("%d", &n);        for (int i = 0; i < n; i++) {            scanf ("%d", &cnt[i]);            for (int j = 1; j <= cnt[i]; j++)                scanf ("%d", &Milk[i][j]);        }        int ans = solve();        printf ("%d %d\n", ans, survice[ans]);    }    return 0;}

Uva10273-eat or not to eat? (Violent)