Problem:
Problem H: Galaxy Wide Web Service

The volume of access to a web service varies from time to time in a day. Also, the hours with the highest
volume of access varies from service to service. For example, a service popular in the United States
may receive more access in the daytime in the United States, while another service popular in Japan may
receive more access in the daytime in Japan. When you develop a web service, you have to design the
system so it can handle all requests made during the busiest hours.

You are a lead engineer in charge of a web service in the 30th century. It’s the era of Galaxy Wide Web
(GWW), thanks to the invention of faster-than-light communication. The service can be accessed from
all over the galaxy. Thus many intelligent creatures, not limited to human beings, can use the service.
Since the volume of access to your service is increasing these days, you have decided to reinforce the
server system. You want to design a new system that handles requests well even during the hours with the
highest volume of access. However, this is not a trivial task. Residents in each planet have their specific
length of a day, say, a cycle of life. The length of a day is not always 24 hours. Therefore, a cycle of the
volume of access are different by planets of users.

You have obtained hourly data of the volume of access for all planets where you provide the service.
Assuming the volume of access follows a daily cycle for each planet, you want to know the highest
volume of access in one hour. It should be a quite easy task for you, a famous talented engineer in the
galaxy.

Input

The input consists of multiple datasets. Each dataset has the following format:

N

d1 t1 q1,0 ... q1,d1-1

...

dN tN qN,0 ... qN,dN-1

N is the number of planets. di (1 ≤ i ≤ N) is the length of a day in the planet i.

ti (0 ≤ ti ≤ di - 1) is the current time of the planet i.

qi, j is the volume of access on the planet i during from the j-th hour to the (j+1)-th hour.

You may assume that N ≤ 100, di ≤ 24, qi, j ≤ 1000000 (1 ≤ i ≤ N, 0 ≤ j ≤ di - 1).

The last dataset is followed by a line containing one zero. This line is not a part of any dataset and should not be processed.

Output

For each dataset, output the maximum volume of access in one hour in a line.

Sample Input

2
4 0 1 2 3 4
2 0 2 1
0

Output for the Sample Input

5

Solution in C (judge verdict: Accepted)

// AOJ 2162: Galaxy Wide Web Service
// 2017.12.13 bal4u@uu

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

#define MAX1 55440
#define MAX2 96577

int a[MAX1+1], b[MAX2+1];
int p1, p2;
int q[25][25];
char f[25], chk[25];
char buf[205], *p;

int getint(void)
{
	int n = 0;
	while (*p >= '0') n = (n<<3) + (n<<1) + (*p++ & 0xf);
	p++;
	return n;
}

int gcd(int a, int b)
{
	int r;
	while (b != 0) r = a % b, a = b, b = r;
	return a;
}

int lcm(int a, int b)
{
	return a/gcd(a, b)*b;
}

int main()
{
	int n, i, j, jj, k, d, t, ans1, ans2;

	chk[13] = chk[17] = chk[19] = chk[23] = 1;

	while (fgets(p=buf, 10, stdin) && *p != '0') {
		n = getint();

		memset(q, 0, sizeof(q));
		memset(f, 1, sizeof(f));

		for (i = 0; i < n; i++) {
			fgets(p=buf, 205, stdin);
			d = getint(), t = getint(); f[d] = 0;
			jj = 0; if (t > 0) jj = d-t; 			
			j = d; while (j--) {
				q[d][jj] += getint();
				if (++jj == d) jj = 0;
			}
		}

		p1 = p2 = 1;
		for (d = 1; d <= 24; d++) {
			if (f[d]) continue;
			if (chk[d]) p2 *= d;
			else p1 = lcm(p1, d);
		}

		memset(a, 0, sizeof(int)*p1);
		memset(b, 0, sizeof(int)*p2);

		for (d = 1; d <= 24; d++) {
			if (f[d]) continue;
			if (chk[d]) {
				for (i = 0; i < d; i++) {
					k = q[d][i];
					for (j = i; j < p2; j += d) b[j] += k;
				}
			} else {
				for (i = 0; i < d; i++) {
					k = q[d][i];
					for (j = i; j < p1; j += d) a[j] += k;
				}
			}
		}

		ans1 = 0; for (i = 0; i < p1; i++) if (a[i] > ans1) ans1 = a[i];
		ans2 = 0; for (i = 0; i < p2; i++) if (b[i] > ans2) ans2 = b[i];
		printf("%d\n", ans1+ans2);
	}
	return 0;
}