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62 lines
1.5 KiB
62 lines
1.5 KiB
#include <bits/stdc++.h>
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using namespace std;
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const int N = 1e5 + 10;
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int a[N], al;
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int b[N], bl;
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void mul(int a[], int &al, int b[], int bl) {
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int c[N] = {0}, cl = al + bl;
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for (int i = 1; i <= al; i++)
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for (int j = 1; j <= bl; j++)
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c[i + j - 1] += a[i] * b[j];
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int t = 0;
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for (int i = 1; i <= al + bl; i++) {
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t += c[i];
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c[i] = t % 10;
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t /= 10;
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}
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memcpy(a, c, sizeof c);
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al = cl;
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//前导0
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while (al > 1 && a[al] == 0) al--;
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}
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//快速幂+高精度 x^k
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void qmi(int x, int k) {
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a[++al] = 1, b[++bl] = x; // 2 ^100 b[1]=2
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while (k) {
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if (k & 1) mul(a, al, b, bl);
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k >>= 1;
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mul(b, bl, b, bl);
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}
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}
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int main() {
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int k = 100; // k=100 2^k
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// 1、利用高精度计算2^p位数 暴力法
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a[++al] = 1, b[++bl] = 2;
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for (int i = 1; i <= k; i++) mul(a, al, b, bl); // O(N)
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printf("%d\n", al);
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for (int i = al; i; i--) printf("%d", a[i]);
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puts("");
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// 2、利用高精度+快速幂来计算2^p位数
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al = 0, bl = 0;
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// memset(a, 0, sizeof a);
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// memset(b, 0, sizeof b);
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// 64 32 16 8 4 2 1
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// 1 1 0 0 1 0 0
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qmi(2, k); // log2(100)=6.6=7
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printf("%d\n", al); // O(log2(N)≈lg(N)=2)
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for (int i = al; i; i--) printf("%d", a[i]);
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puts("");
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// 3、利用对数公式计算2^k位数
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// k*log10(2)+1
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printf("%d\n", int(100 * log10(2) + 1)); // O(1)
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return 0;
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} |