BZOJ5300 [Cqoi2018]九连环 【dp + 高精】

题目链接

BZOJ5300

题解

这题真的是很丧病，，卡高精卡到哭
我们设\(f[i]\)表示卸掉前\(i\)个环需要的步数
那么

\[f[i] = 2*f[i - 2] + f[i - 1] + 1 \]

直接高精递推不仅\(MLE\)而且\(TLE\)
然后就要用到数学求通项公式，或者打表找规律

\[f[n] = \lfloor \frac{2^{n + 1}}{3} \rfloor \]

高精乘低精就可以过了

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 100005,B = 100000000,maxm = 100005,INF = 1000000000;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
struct NUM{
	LL s[100000],len;
	NUM(){cls(s); len = 0;}
	void out(){
		if (!len){putchar('0'); return;}
		printf("%lld",s[len - 1]);
		for (int i = len - 2; i >= 0; i--)
			printf("%08lld",s[i]);
	}
}F;
inline void operator *=(NUM& a,const int& b){
	LL tmp,carry = 0;
	for (int i = 0; i < a.len; i++){
		tmp = 1ll * a.s[i] * b + carry;
		a.s[i] = tmp % B;
		carry = tmp / B;
	}
	while (carry) a.s[a.len++] = carry % B,carry /= B;
}
inline NUM operator /(const NUM& a,const int& b){
	NUM c;
	c.len = a.len;
	LL tmp,carry = 0;
	for (int i = a.len - 1; i >= 0; i--){
		tmp = a.s[i] + 1ll * carry * B;
		if (tmp < b) c.s[i] = 0,carry = tmp;
		else c.s[i] = tmp / b,carry = tmp % b;
	}
	//if (carry) c = c + 1;
	while (c.len && !c.s[c.len - 1]) c.len--;
	return c;
}
int main(){
	int T = read();
	while (T--){
		int n = read() + 1,bin = 1 << 30;
		F.len = 1; F.s[0] = 1;
		int tmp = n / 30,t = n % 30;
		while (tmp--) F *= bin;
		while (t--) F *= 2;
		F = F / 3;
		F.out(); puts("");
	}
	return 0;
}