UVA 10689 - Yet another Number Sequence

题目链接

题意：斐波那契给前两项，求出第n项，并保留m位

思路：挺裸的矩阵快速幂，就是取模的值是10^m

代码：

#include <cstdio>
#include <cstring>

const int mod[5] = {0, 10, 100, 1000, 10000};
int t, a, b, n, m;

struct mat {
	int v[2][2];
	mat() {memset(v, 0, sizeof(v));}
	mat operator * (mat c) {
		mat ans;
		for (int i = 0; i < 2; i++) {
			for (int j = 0; j < 2; j++) {
				for (int k = 0; k < 2; k++) {
					ans.v[i][j] = (ans.v[i][j] + v[i][k] * c.v[k][j]) % mod[m];
				}
			}
		}
		return ans;
	}
};

mat pow_mod(mat x, int k) {
	mat ans;
	ans.v[0][0] = ans.v[1][1] = 1;
	while (k) {
		if (k&1) ans = ans * x;
		x = x * x;
		k >>= 1;
	}
	return ans;
}

int solve() {
	if (n == 0) return a;
	if (n == 1) return b;
	mat ans;
	ans.v[0][0] = ans.v[0][1] = ans.v[1][0] = 1;
	ans = pow_mod(ans, n - 1);
	return (ans.v[0][0] * b + ans.v[0][1] * a);
}

int main() {
	scanf("%d", &t);
	while (t--) {
		scanf("%d%d%d%d", &a, &b, &n, &m);
		printf("%d\n", solve() % mod[m]);
	}
	return 0;
}