fzu 1911 Construct a Matrix(矩阵快速幂+规律）

题目链接：fzu 1911 Construct a Matrix

题目大意：给出n和m，f[i]为斐波那契数列，s[i]为斐波那契数列前i项的和。r = s
% m。构造一个r * r的矩阵，只能使用-1、0、1。使得矩阵的每行每列的和都不相同，输出方案，不行的话输出No。

解题思路：求r的话用矩阵快速幂求，每次模掉m，

{ {1， 1， 0}， {1， 0， 0}， {1， 1， 1} } * { f[i], f[i -1], s[i] } = { f[i + 1], f[i], s[i + 1] }.

然后求出r后，若r是奇数或0，则矩阵不存在；r为偶数时，只要按照规律建立矩阵就可以了。

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

const int M = 10;
const int N = 205;

int n, m, r;

struct Mul {
int s[M][M];
Mul() { memset(s, 0, sizeof(s)); }
Mul operator * (const Mul& c) {
Mul ans;

for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
ans.s[i][j] = 0;
for (int k = 0; k < 3; k++)
ans.s[i][j] = (ans.s[i][j] + s[i][k] * c.s[k][j] ) % m;
}
}
return ans;
}

void put() {
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++)
printf("%d ", s[i][j]);
printf("\n");
}
}
};

Mul MulPow(Mul a, int t) {
if (t == 1) return a;

Mul x = MulPow(a, t / 2);

x = x * x;

if (t % 2) x = x * a;

return x;
}

void init() {
if (n > 2) {
Mul a;
a.s[0][0] = a.s[0][1] = a.s[1][0] = a.s[2][0] = a.s[2][1] = a.s[2][2] = 1;

Mul ans = MulPow(a, n - 2);

r = (ans.s[2][0] + ans.s[2][1] + ans.s[2][2] * 2) % m;
} else if (n == 2) {
r = 2 % m;
} else if (n == 1) {
r = 1;
}
}

void solve() {
if (r == 0 || r % 2)
printf("No\n");
else {
int v

;
memset(v, -1, sizeof(v));
printf("Yes\n");

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

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

int main () {
int cas;
scanf("%d", &cas);
for (int i = 1; i <= cas; i++) {
scanf("%d%d", &n, &m);
printf("Case %d: ", i);

init();

solve();
}
return 0;
}