Problem:
Score : 200 points

Problem Statement

There are N pieces of source code. The characteristics of the i-th code is represented by M integers A_{i1}, A_{i2}, ..., A_{iM}.

Additionally, you are given integers B_1, B_2, ..., B_M and C.

The i-th code correctly solves this problem if and only if A_{i1} B_1 + A_{i2} B_2 + ... + A_{iM} B_M + C > 0.

Among the N codes, find the number of codes that correctly solve this problem.

Constraints

All values in input are integers.

1 \leq N, M \leq 20

-100 \leq A_{ij} \leq 100

-100 \leq B_i \leq 100

-100 \leq C \leq 100

Input

Input is given from Standard Input in the following format:

N M C
B_1 B_2 ... B_M
A_{11} A_{12} ... A_{1M}
A_{21} A_{22} ... A_{2M}
\vdots
A_{N1} A_{N2} ... A_{NM}

Output

Print the number of codes among the given N codes that correctly solve this problem.

Sample Input 1

2 3 -10
1 2 3
3 2 1
1 2 2

Sample Output 1

1

Only the second code correctly solves this problem, as follows:

Since 3 \times 1 + 2 \times 2 + 1 \times 3 + (-10) = 0 \leq 0, the first code does not solve this problem.

1 \times 1 + 2 \times 2 + 2 \times 3 + (-10) = 1 > 0, the second code solves this problem.

Sample Input 2

5 2 -4
-2 5
100 41
100 40
-3 0
-6 -2
18 -13

Sample Output 2

2

Sample Input 3

3 3 0
100 -100 0
0 100 100
100 100 100
-100 100 100

Sample Output 3

0

All of them are Wrong Answer. Except yours.

Solution in C (Clang 3.8.0) (judge verdict: Accepted)

#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <inttypes.h>
#include <ctype.h>
#include <stdint.h>
#include <string.h>
#include <wchar.h>

#define N_MAX (100)
#define P_MAX (100)
#define DP_ARRAY_SIZE (N_MAX * P_MAX / 32 + 1)

#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define ABS(a) ((a) < 0 ? -(a) : (a))
#define ABSS(a, b) ((a) > (b) ? (a) - (b) : (b) - (a))

int compare_f(const void *a, const void *b) {
	return *((size_t *)a) < *((size_t *)b) ? -1 : 1;
}

static size_t comb(const size_t n, const size_t r) {
	size_t result = 1;

	for (size_t i = 0; i < r; i++) {
		result *= n - i;
		result /= i + 1;
	}

	return result;
}

static uint64_t gcd(uint64_t m, uint64_t n)
{
	uint64_t temp;

	while (m%n != 0) {
		temp = n;
		n = m % n;
		m = temp;
	}

	return n;
}

int main(void) {
	size_t N, M;
	int32_t B[20];
	int32_t C;

	scanf("%zu %zu %"PRId32"\n", &N, &M, &C);
	for (size_t i = 0; i < M; i++) {
		scanf("%"PRId32, &B[i]);
		if (i < M - 1) {
			scanf(" ");
		}
		else {
			scanf("\n");
		}
	}

	size_t count = 0;

	for (size_t i = 0; i < N; i++) {
		int32_t sum = C;
		for (size_t j = 0; j < M; j++) {
			int32_t A;
			scanf("%"PRId32, &A);
			if (j < M - 1) {
				scanf(" ");
			}
			else {
				if (i < N - 1) {
					scanf("\n");
				}
			}
			sum += B[j] * A;
		}

		if (0 < sum) {
			count++;
		}
	}

	printf("%zu\n", count);

	return 0;
}