fzu 1911 C. Construct a Matrix

C. Construct a Matrix

Time Limit: 1000ms
Case Time Limit: 1000ms
Memory Limit: 32768KB
Special Judge

64-bit integer IO format: %I64d Java class name: Main
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There is a set of matrixes that are constructed subject to the following constraints:
1. The matrix is a S(n)×S(n) matrix;

2. S(n) is the sum of the first n Fibonacci numbers modulus m, that is S(n) = (F1 + F2 + … + Fn) % m;

3. The matrix contains only three kinds of integers ‘0’, ‘1’ or ‘-1’;

4. The sum of each row and each column in the matrix are all different.

Here, the Fibonacci numbers are the numbers in the following sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …

By definition, the first two Fibonacci numbers are 1 and 1, and each remaining number is the sum of the previous two.

In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation Fn = Fn-1 + Fn-2, with seed values F1 = F2 = 1.

Given two integers n and m, your task is to construct the matrix.

Input

The first line of the input contains an integer T (T <= 25), indicating the number of cases. Each case begins with a line containing two integers n and m (2 <= n <= 1,000,000,000, 2 <= m <= 200).

Output

For each test case, print a line containing the test case number (beginning with 1) and whether we could construct the matrix. If we could construct the matrix, please output “Yes”, otherwise output “No” instead. If there are multiple solutions, any one is accepted and then output the S(n)×S(n) matrix, separate each integer with an blank space (as the format in sample).

Sample Input

2
2 3
5 2

Sample Output

Case 1: Yes
-1 1
0 1
Case 2: No

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题意:求fib数列前n项和，以这个和%m作为矩阵的r，然后构造矩阵满足所有值为0，1 或 -1，并且每行每列和都不相等。

思路:第一步矩阵快速幂，第二步找规律。

代码:

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

const int N = 205;
int t, n, m, r;

struct mat {
int v[3][3];
mat() {
memset(v, 0, sizeof(v));
}
mat operator * (mat &b) {
mat c;
for (int i = 0; i < 3; i ++)
for (int j = 0; j < 3; j ++)
for (int k = 0; k < 3; k ++)
c.v[i][j] = (v[i][k] * b.v[k][j] + c.v[i][j]) % m;
return c;
}
};

mat pow_mod(mat a, int k) {
if (k == 1 || k == 0)
return a;
mat c = pow_mod(a * a, k / 2);
if (k & 1)
c = c * a;
return c;
}

void init() {
scanf("%d%d", &n, &m);
mat start;
start.v[0][0] = start.v[0][1] = start.v[1][0] = start.v[2][0] = start.v[2][1] = start.v[2][2] = 1;
if (n == 1)
r = 1;
else if (n == 2)
r = 2;
else {
mat end = pow_mod(start, n - 2);
r = (end.v[2][0] + end.v[2][1] + end.v[2][2] * 2) % m;
}
}

void solve() {
int s

;
memset(s, -1, sizeof(s));

if (r == 0 || r % 2)
printf("No\n");
else {
printf("Yes\n");
for (int i = 1; i <= r; i++) {

if (i % 2) {
int tmp = r / 2 + (i + 1) / 2;
s[tmp][i] = 0;
for (int j = tmp + 1; j <= r; j++)
s[j][i] = 1;
} else {
int tmp = (r - i) / 2;
for (int j = tmp + 1; j <= r; j++)
s[j][i] = 1;
}
}

for (int i = 1; i <= r; i++) {
//   int sum = 0;
for (int j = 1; j < r; j++) {
printf("%d ", s[i][j]);
//	sum += s[i][j];
}
printf("%d\n", s[i][r]);
}

/*
for (int j = 1; j <= r; j++) {
int sum = 0;
for (int i = 1; i <= r; i++)
sum += s[i][j];
printf("%d ", sum);
}
printf("\n");
*/
}
}

int main() {
int cas = 0;
scanf("%d", &t);
while (t --) {
init();
printf("Case %d: ", ++cas);
solve();
}
return 0;
}