54 lines
1.2 KiB
C++
54 lines
1.2 KiB
C++
#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 = min(500, 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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//计算 2^p-1的值
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int p;
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cin >> p;
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//利用快速幂，计算2^p
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qmi(2, p);
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//最后一位减去一个1，因为2^p最后一位肯定不是0，所以减1不会产生借位，直接减去即可！
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a[1]--;
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//一共多少位
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printf("%d\n", (int)(p * log10(2) + 1));
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for (int i = 500; i; i--) {
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printf("%d", a[i]);
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//该换行了,就是到了第二行的行首
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if ((i - 1) % 50 == 0) puts("");
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}
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return 0;
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} |