C. Construct a Matrix
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
Output
Sample Input
2
2 3
5 2
Sample Output
Case 1: Yes
-1 1
0 1
Case 2: No
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[N][N];
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;
}