题意:f(n) = a1f(n − 1) + a2f(n − 2) + a3f(n − 3) + . . . + adf(n − d), for n > d,求f (n) % m。训练指南的题目
分析:令:,
.则
#include <bits/stdc++.h>
int d, n, m;
int a[16], f[16];
struct Mat {
int m[17][17];
int row, col;
Mat() {
//row = col = 16;
memset (m, 0, sizeof (m));
}
void init(int sz) {
row = col = sz;
for (int i=1; i<row; ++i) {
m[i][i+1] = 1;
}
int c = sz - 1;
for (int i=2; i<=col; ++i) {
m[sz][i] = a[c--];
}
}
void change(int sz) {
row = col = sz;
for (int i=1; i<=sz; ++i) {
m[i][i] = 1;
}
}
};
Mat operator * (const Mat &a, const Mat &b) {
Mat ret;
ret.row = a.row; ret.col = b.col;
for (int i=1; i<=a.row; ++i) {
for (int j=1; j<=b.col; ++j) {
for (int k=1; k<=a.col; ++k) {
int &r = ret.m[i][j];
r = (r + 1ll * a.m[i][k] * b.m[k][j]) % m;
}
}
}
return ret;
}
Mat operator ^ (Mat x, int n) {
Mat ret; ret.change (d+1);
while (n) {
if (n & 1) {
ret = ret * x;
}
x = x * x; n >>= 1;
}
return ret;
}
//Running_Time
int main() {
while (scanf ("%d%d%d", &d, &n, &m) == 3) {
if (!d && !n && !m) {
break;
}
for (int i=1; i<=d; ++i) {
scanf ("%d", a+i);
}
for (int i=1; i<=d; ++i) {
scanf ("%d", f+i);
}
if (n <= d) {
printf ("%d\n", f[n] % m);
} else {
Mat ans, Fd;
ans.init (d + 1);
ans = ans ^ (n - d);
Fd.row = d + 1; Fd.col = 1;
for (int i=2; i<=d+1; ++i) {
Fd.m[i][1] = f[i-1];
}
ans = ans * Fd;
printf ("%d\n", ans.m[d+1][1]);
}
}
return 0;
}