_bzoj1002 [FJOI2007]轮状病毒【瞎搞】

传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=1002

这种题真是有毒,很多叼一点的都用matrix tree定理推出了递推公式,也有一些用好几维递推也可以解,然而我一个都看不懂。总结:下次遇到这种题就直接打表找规律,这是坠吼滴!

#include <cstdio>
#include <cstring>

const int maxn = 105, mod = 10000;

int n;
struct gao {
	int data[100], wei;
	gao(void) {
		memset(data, 0, sizeof data);
		wei = 0;
	}
	gao operator+(int aa) const {
		gao rt;
		rt.data[0] = aa;
		for (int i = 0; i < wei; ++i) {
			rt.data[i] += data[i];
			if (rt.data[i] >= mod) {
				rt.data[i] -= mod;
				++rt.data[i + 1];
			}
		}
		rt.wei = wei + (rt.data[wei]? 1: 0);
		return rt;
	}
	gao operator-(const gao & rhs) const {
		gao rt;
		for (int i = 0; i < wei; ++i) {
			rt.data[i] += data[i] - rhs.data[i];
			if (rt.data[i] < 0) {
				rt.data[i] += mod;
				--rt.data[i + 1];
			}
		}
		rt.wei = wei;
		while (rt.wei > 0 && !rt.data[rt.wei - 1]) {
			--rt.wei;
		}
		return rt;
	}
	gao operator*(int aa) const {
		gao rt;
		for (int i = 0; i < wei; ++i) {
			rt.data[i] += data[i] * aa;
			rt.data[i + 1] += rt.data[i] / mod;
			rt.data[i] %= mod;
		}
		rt.wei = wei + (rt.data[wei]? 1: 0);
		return rt;
	}
	void print(void) {
		printf("%d", data[wei - 1]);
		for (int i = wei - 2; i >= 0; --i) {
			printf("%04d", data[i]);
		}
	}
} f[maxn];

int main(void) {
	f[1].data[0] = 1;
	f[1].wei = 1;
	f[2].data[0] = 5;
	f[2].wei = 1;
	scanf("%d", &n);
	for (int i = 3; i <= n; ++i) {
		f[i] = f[i - 1] * 3 - f[i - 2] + 2;
	}
	f[n].print();
	puts("");
	return 0;
}

  

posted @ 2016-12-13 20:53  ciao_sora  阅读(160)  评论(0编辑  收藏  举报