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发表于 2023-9-24 12:17:48
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显示全部楼层
一个高精度幂的模板,输入两个数字,然后按回车。
- #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 = 133015;
- 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)
- 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);
- string a;
- int b;
- while(cin>>a>>b) cout<<Pow(a,b)<<endl;
- return 0;
- }
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