算法设计与分析入门学习练习之二

//计算1/1!-1/3!+1/5!-1/7!+…+(-1)^(n+1)/(2n-1)!
float mathTest(int n){

    int sign = 1;
    float sum = 1, t = 1;
    for (int i = 1; i <= n - 1; i++)
    {
        sign = sign*(-1);
        t = t*(2 * i)*(2 * i + 1);
        sum = sum + sign / t;
        
    }
    return sum;
}
//判断一个是否为完数(如：28=1+2+4+7+14)
//寻找(0，n]以内的所有完数
void IsPnum(int n){
    int i, a[100];
    for (i = 1; i <= n;i++)
    {
        int s = 1, k = 0;
        for (int j = 2; j < i;j++)
        {
            if ((i%j)==0)
            {
                s = s + j;
                a[k] = j;
                k++;
            }
        }
        if ((i==s)&&(s!=1))
        {
            cout << s << ",it‘s factors are:" << "1";
            for (int i = 0; i < k;i++)
            {
                cout << "," << a[i];
            }
            cout << endl;
        }
    }
    
}
//求矩阵的鞍点，即行上最小而列上最大的点
void FindAnDian(){

    //输入一个n*n的矩阵
    cout << "input n*n matric: " << endl;
    int a[3][3], n = 3, col, row ,AD=0;
    //cin >> n;
    for (int i = 0; i < n;i++)
    {
        for (int j = 0; j < n;j++)
        {
            cin >> a[i][j];
        }
    }
    //打印矩阵
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            cout<< a[i][j];
        }
        cout << endl;
    }
    //寻找鞍点
    for (int i = 0; i < n;i++)
    {
        int t = a[i][0];
        for (int j = 0; j < n;j++)
        {
            if (t>a[i][j])
            {
                t = a[i][j];
                col = j;
            }
        }
        for (row = 0; row < n;row++)
        {
            if (t<a[row][col])
            {
                break;//如果第col列中有元素大于t;则直接终止判断是否为列最大元素
            }
        }
        if (row<n)
        {
            continue;//如果row<n即证明第i行的最小元素不是第col列的最大元素直接结束本次循环。寻找下一行元素的最小值。
        }
        else
        {
            cout << "the result is a[" << i << "]" << "[" << col << "]" << endl;
            AD = 1;
            break;
        }


    }
    if (AD==0)
    {
        cout << "Non result" << endl;
    }
}