若有int n=10，m=20，*p=&n，*q=&m，则以下赋值话句中错误的是a，p=q,b,* p=*q, p=*q

根据您提供的信息，以下赋值语句中错误的是 `p=*q`。因为 `p` 是一个指向整型变量 `n` 的指针，而 `*q` 是整型变量 `m` 的值，类型不匹配，不能将一个整型值直接赋给一个指针变量。另外，`a` 和 `b` 都是表达式，没有被完整地给出，无法判断它们是否正确。因此，正确的代码应该是：

int n = 10;
int m = 20;
int* p = &n;
int* q = &m;
*p = *q;  // 将 m 的值赋给 n，即将 20 赋给变量 n

这样就可以将变量 `m` 的值赋给变量 `n`。

相关问题

#include <iostream> #include <cstdlib> using namespace std; int matrixChainOrder(int p[], int n, int** m, int** s) { for(int i = 1; i <= n; i++) { m[i][i] = 0; } for(int l = 2; l <= n; l++) { for(int i = 1; i <= n-l+1; i++) { int j = i+l-1; m[i][j] = INT_MAX; for(int k = i; k <= j-1; k++) { int q = m[i][k] + m[k+1][j] + p[i-1]*p[k]*p[j]; if(q < m[i][j]) { m[i][j] = q; s[i][j] = k; } } } } return m[1][n]; } int main() { int n = 3; int p[] = {8, 1, 8}; int** m = new int*[n+1]; int** s = new int*[n+1]; for(int i = 1; i <= n; i++) { m[i] = new int[n+1]; s[i] = new int[n+1]; } int minMultiplications = matrixChainOrder(p, n, m, s); cout << "The minimum number of multiplications is: " << minMultiplications << endl; cout << "The solutions to the subproblems are:" << endl; for(int i = 1; i <= n; i++) { for(int j = 1; j <= n; j++) { cout << "(" << m[i][j] << ", " << s[i][j] << ") "; } cout << endl; } for(int i = 1; i <= n; i++) { delete[] m[i]; delete[] s[i]; } delete[] m; delete[] s; return 0; }请在此代码基础上稍加修改，避免输出的矩阵元素值异常大

这段代码存在一个问题，就是在未初始化数组 `m` 和 `s` 的值时，直接对其进行赋值操作，在这种情况下，它们的值是不确定的，可能会导致输出异常大的情况。因此，我们需要在对数组进行赋值操作之前，先将它们的值初始化为 0。

修改后的代码如下所示：

#include <iostream>
#include <cstdlib>
using namespace std;

int matrixChainOrder(int p[], int n, int** m, int** s) {
    // 初始化数组 m 和 s
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= n; j++) {
            m[i][j] = 0;
            s[i][j] = 0;
        }
    }

    for(int i = 1; i <= n; i++) {
        m[i][i] = 0;
    }
    for(int l = 2; l <= n; l++) {
        for(int i = 1; i <= n-l+1; i++) {
            int j = i+l-1;
            m[i][j] = INT_MAX;
            for(int k = i; k <= j-1; k++) {
                int q = m[i][k] + m[k+1][j] + p[i-1]*p[k]*p[j];
                if(q < m[i][j]) {
                    m[i][j] = q;
                    s[i][j] = k;
                }
            }
        }
    }
    return m[1][n];
}

int main() {
    int n = 3;
    int p[] = {8, 1, 8};
    int** m = new int*[n+1];
    int** s = new int*[n+1];
    for(int i = 1; i <= n; i++) {
        m[i] = new int[n+1];
        s[i] = new int[n+1];
    }

    int minMultiplications = matrixChainOrder(p, n, m, s);
    cout << "The minimum number of multiplications is: " << minMultiplications << endl;
    cout << "The solutions to the subproblems are:" << endl;
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= n; j++) {
            cout << "(" << m[i][j] << ", " << s[i][j] << ") ";
        }
        cout << endl;
    }
    for(int i = 1; i <= n; i++) {
        delete[] m[i];
        delete[] s[i];
    }
    delete[] m;
    delete[] s;
    return 0;
}

修改后的代码会对数组 `m` 和 `s` 进行初始化，初始化后再进行赋值操作，从而避免输出异常大的矩阵元素值。

from Crypto.Util.number import * import gmpy2 def get_params(): a,b = [getPrime(128) for _ in range(2)] a,b = [bin(i)[2:].zfill(128) for i in [a,b]] #使用bin(i)将其转换为二进制字符串，并使用[2:]来去除开头的"0b"。（bin(i)返回的字符串形式以"0b"开头，表示该字符串是二进制表示形式） p = int((a + b),2) #上一行已经改为了二进制，那么在这里拼接后改为十进制，并且这里是运算，不会改变ab的值，拼接更不会对其改变 q = int((b + a),2) a, b = int(a, 2), int(b, 2) #将二进制字符串a和b分别转换为十进制整数，并将它们分别赋值给变量a和b。 return gmpy2.next_prime(p),gmpy2.next_prime(q),a,b flag = b'flag{******}' m = bytes_to_long(flag) p,q,a,b = get_params() n = p * q stream = [getPrime(128)] for i in range(114): num = (a * stream[-1] + b) % p stream.append(num) e = gmpy2.next_prime(stream[1] * stream[14] + stream[51] * stream[4]) c = pow(m,e,n) print(f'n = {n}') print(f'c = {c}') # n = 6983103504667653623549839927036982238702989760981440695828662474048376735590137382537887815211794982723527122829225808557810219686458488017050037179654123 # c = 6796222297027759937096057530534388748539384795006307524526800462652134886498275107011127280721467103062302906430290388832305421496018525958278394725734527 请写出解密脚本

解密脚本如下：

from Crypto.Util.number import long_to_bytes
import gmpy2

def get_params():
    a,b = [getPrime(128) for _ in range(2)]
    a,b = [bin(i)[2:].zfill(128) for i in [a,b]]
    p = int((a + b),2)
    q = int((b + a),2)
    a, b = int(a, 2), int(b, 2)
    return gmpy2.next_prime(p),gmpy2.next_prime(q),a,b

def decrypt(c, p, q, a, b):
    phi_n = (p - 1) * (q - 1)
    d = gmpy2.invert(e, phi_n)
    m = pow(c, d, n)
    return long_to_bytes(m)

n = 6983103504667653623549839927036982238702989760981440695828662474048376735590137382537887815211794982723527122829225808557810219686458488017050037179654123
c = 6796222297027759937096057530534388748539384795006307524526800462652134886498275107011127280721467103062302906430290388832305421496018525958278394725734527
e = gmpy2.next_prime(stream[1] * stream[14] + stream[51] * stream[4])
p,q,a,b = get_params()

plaintext = decrypt(c, p, q, a, b)
print(f'Plaintext: {plaintext.decode()}')

请注意，解密脚本中的 `stream` 数组在给定的代码中没有明确定义，因此需要根据上下文做适当的调整。此外，需要确保 `gmpy2` 库已正确安装。