题解 P1760 【通天之汉诺塔】
Shikita
2018-10-06 15:39:46
# 高精度幂方
想必题目非常好懂吧，就是2^n-1，~~看一眼样例那熟悉的数字猜都猜出来了吧~~
但是再看一眼数据 n<=15000，很明显就是一道高精度了
其实代码都差不多，别的都是80+行，还有几个~~无耻的~~pyhon两行我就不说了
但是现在NOI/NOIP还未支持pyhon，所以高精度还是要打的嘛
而且这题最无耻的是，第一个点是0！！！！
我因此疯狂MLE，惊了！
由于递归调用，所以输出那里就丑陋了一点，请大佬们谅解
剧透：FFT优化
（其实我就是来宣传FFT优化高精度幂单调）
## 代码
```
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <map>
#include <queue>
#include <set>
#include <vector>
using namespace std;
#define L(x) (1 << (x))
const double PI = acos(-1.0);
const int Maxn = 80000;
double ax[Maxn], ay[Maxn], bx[Maxn], by[Maxn];
char sa[Maxn/2],sb[Maxn/2];
int sum[Maxn];
int x1[Maxn],x2[Maxn];
int revv(int x, int bits)
{
int ret = 0;
for (int i = 0; i < bits; i++)
{
ret <<= 1;
ret |= x & 1;
x >>= 1;
}
return ret;
}
void fft(double * a, double * b, int n, bool rev)
{
int bits = 0;
while (1 << bits < n) ++bits;
for (int i = 0; i < n; i++)
{
int j = revv(i, bits);
if (i < j)
swap(a[i], a[j]), swap(b[i], b[j]);
}
for (int len = 2; len <= n; len <<= 1)
{
int half = len >> 1;
double wmx = cos(2 * PI / len), wmy = sin(2 * PI / len);
if (rev) wmy = -wmy;
for (int i = 0; i < n; i += len)
{
double wx = 1, wy = 0;
for (int j = 0; j < half; j++)
{
double cx = a[i + j], cy = b[i + j];
double dx = a[i + j + half], dy = b[i + j + half];
double ex = dx * wx - dy * wy, ey = dx * wy + dy * wx;
a[i + j] = cx + ex, b[i + j] = cy + ey;
a[i + j + half] = cx - ex, b[i + j + half] = cy - ey;
double wnx = wx * wmx - wy * wmy, wny = wx * wmy + wy * wmx;
wx = wnx, wy = wny;
}
}
}
if (rev)
{
for (int i = 0; i < n; i++)
a[i] /= n, b[i] /= n;
}
}
int solve(int a[],int na,int b[],int nb,int ans[])
{
int len = max(na, nb), ln;
for(ln=0; L(ln)<len; ++ln);
len=L(++ln);
for (int i = 0; i < len ; ++i)
{
if (i >= na) ax[i] = 0, ay[i] =0;
else ax[i] = a[i], ay[i] = 0;
}
fft(ax, ay, len, 0);
for (int i = 0; i < len; ++i)
{
if (i >= nb) bx[i] = 0, by[i] = 0;
else bx[i] = b[i], by[i] = 0;
}
fft(bx, by, len, 0);
for (int i = 0; i < len; ++i)
{
double cx = ax[i] * bx[i] - ay[i] * by[i];
double cy = ax[i] * by[i] + ay[i] * bx[i];
ax[i] = cx, ay[i] = cy;
}
fft(ax, ay, len, 1);
for (int i = 0; i < len; ++i)
ans[i] = (int)(ax[i] + 0.5);
return len;
}
string mul(string sa,string sb)
{
int l1,l2,l;
int i;
string ans;
memset(sum, 0, sizeof(sum));
l1 = sa.size();
l2 = sb.size();
for(i = 0; i < l1; i++)
x1[i] = sa[l1 - i - 1]-'0';
for(i = 0; i < l2; i++)
x2[i] = sb[l2-i-1]-'0';
l = solve(x1, l1, x2, l2, sum);
for(i = 0; i<l || sum[i] >= 10; i++)
{
sum[i + 1] += sum[i] / 10;
sum[i] %= 10;
}
l = i;
while(sum[l] <= 0 && l>0) l--;
for(i = l; i >= 0; i--) ans+=sum[i] + '0';
return ans;
}
string Pow(string a,int n)
{
if(n==1) return a;
if(n&1) return mul(Pow(a,n-1),a);
string ans=Pow(a,n/2);
return mul(ans,ans);
}
int main()
{
cin.sync_with_stdio(false);//没什么用的东西
int n;
string s="2";
cin>>n;
if(n==0) {cout<<0<<endl;return 0;}
string ss=Pow(s,n);
int x=ss.size();
ss[x-1]=(ss[x-1]-'0'-1)+'0';
cout<<ss<<endl;
return 0;
}
```