题目链接
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}
floor
]
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;
}