算法设计作业

def quicksort(arr):
    if len(arr) <= 1:
        return arr
    pivot = arr[len(arr) // 2]
    left = [x for x in arr if x < pivot]
    middle = [x for x in arr if x == pivot]
    right = [x for x in arr if x > pivot]
    return quicksort(left) + middle + quicksort(right)


# 测试代码
n = int(input())
arr = []
for i in range(n):
    x = int(input())
    arr.append(x)
arr = quicksort(arr)
for index,a in enumerate(arr):
    if index == len(arr) - 1:
        print(f'{a}')
        break
    print(f'{a}', end=' ')


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#include
#include
int a[110];
int n;
void merge(int a[],int tem[],int p,int q,int r)
{
    int i=p;int j=q+1;int pos=p;
    while(i<=q&&j<=r){
        if(a[i]

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7-10 c++

#include
using namespace std;
const int N = 1010;
int f[N];
int v[N], w[N];
int n, m;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    cin >> m >> n;
    for (int i = 1; i <= n; i ++ ) cin >> v[i] >> w[i];

    for (int i = 1; i <= n; i++) {
        for (int j = m; j >= v[i]; j--) {
            f[j] = max(f[j], f[j - v[i]] + w[i]);
        }
    }

    cout << f[m] << endl;
}
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7-11 python

def matrix_chain_order(p):
    n = len(p) - 1 
    m = [[0 for x in range(n + 1)] for y in range(n + 1)]


    for L in range(2, n + 1):
        for i in range(1, n - L + 2):
            j = i + L - 1
            m[i][j] = float('inf')
            for k in range(i, j):
                q = m[i][k] + m[k + 1][j] + p[i - 1] * p[k] * p[j]
                if q < m[i][j]:
                    m[i][j] = q

    return m[1][n]


if __name__ == "__main__":
    n = int(input())
    dims = list(map(int, input().split()))
    print(matrix_chain_order(dims))


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7-12 c++

#include 
#include 
#include 
using namespace std;
void LCS(string s1,string s2)
{
    int m=s1.length()+1;
    int n=s2.length()+1;
    int **c;
    int **b;
    c=new int* [m];
    b=new int* [m];
    for(int i=0;i=c[i+1][j])
            {
                c[i+1][j+1]=c[i][j+1];
                b[i+1][j+1]=2;          //2表示箭头向  上
            }
            else
            {
                c[i+1][j+1]=c[i+1][j];
                b[i+1][j+1]=3;          //3表示箭头向  左
            }
        }
    }
    stack same;                   //存LCS字符
    stack same1,same2;             //存LCS字符在字符串1和字符串2中对应的下标，方便显示出来
    for(int i = m-1,j = n-1;i >= 0 && j >= 0; )
    {
        if(b[i][j] == 1)
        {
            i--;
            j--;
            same.push(s1[i]);
            same1.push(i);
            same2.push(j);
        }
        else if(b[i][j] == 2)
                i--;
             else
                j--;
    }
    for(int i=0;i