随手练——麦森数（高精度快速幂）

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <vector>
#include <algorithm>
using namespace std;

int bits;

void carry(vector<int>& a) {
    for (int i = 0; i < a.size(); i++) {
        if (i != a.size() - 1)a[i + 1] += a[i] / 10;
        else if (a[i] >= 10)a.push_back(a[i] / 10);
        a[i] %= 10;
    }
    while (a.size() > 500) {
        a.pop_back();
        //bits++;
    }
}

void multi(vector<int>& a, vector<int>b) {
    int k = 0;
    vector<int>res(a.size());
    for (int i = 0; i < b.size(); i++) {
        for (int j = 0; j < a.size(); j++) {
            if (k + j >= res.size())
                res.push_back(0);
            res[k + j] += a[j] * b[i];
        }
        k++;
    }
    carry(res);

    a = res;
}

void quick_power(vector<int>& a, int k) {
    vector<int>res;
    res.push_back(1);
    while (k) {
        if (k & 1)
            multi(res, a);
        k >>= 1;
        if(k)
        multi(a, a);
    }
    a = res;
}

int main() {
    int p, k = 1;
    cin >> p;
    vector<int>a, b;
    a.push_back(2); 
    cout << (int)(log10(2) * p + 1) << endl;
    quick_power(a, p);
    
    if (a.size() < 500) {
        for (int i = a.size(); i < 500; i++, k++) {
            cout << 0;
            if (k % 50 == 0)cout << endl;
        }
    }
    a[0] -= 1;
    
    for (int i = a.size() - 1; i >= 0; i--, k++) {
        cout << a[i];
        if (k % 50 == 0)cout << endl;
    }
    cout << endl;
    return 0;
}

#include <iostream>
#include <string.h>
#include <math.h>

using namespace std;

int a[1001] = { 2 }, b[1001], res[1001], temp[1001];

void carry(int *arr) {
    for (int i = 0; i < 500; i++) {
        arr[i + 1] += arr[i] / 10;
        arr[i] %= 10;
    }
}

void multi(int* a, int* b, int* res) {
    memset(temp, 0, sizeof(temp));
    for (int i = 0; i < 500; i++) {
        for (int j = 0; j < 500; j++) {
            temp[i + j] += a[j] * b[i];
        }
    }
    carry(temp);
    memcpy(res, temp, sizeof(temp));
}

void quick_power(int k) {
    res[0] = 1;
    while (k) {
        if (k & 1)
            multi(res, a, res);
        k >>= 1;
        if (k)
            multi(a, a, a);
    }
}

int main() {
    int p, k = 1;
    cin >> p;
    int bits = (int)(log10(2) * p + 1);
    cout << bits << endl;
    quick_power(p);
    //2^n次方，不可能为0，所以可以直接减1
    res[0] -= 1;
    for (int i = 499; i >= 0; i--, k++) {
        cout << res[i];
        if (k % 50 == 0)cout << endl;
    }
    cout << endl;
    return 0;
}