暴力求解法——简单枚举

Posted 2022-02-23 小乖乖的臭坏坏

例题1：

代码：

public class uva725 
    public static void main(String[] args) 
        int n = 62;
        for (int a=12345;a<=98765;a++)
            if(a%n!=0)
                continue;
            
            else 
                int b = a / n;
                if(checkNum(a, b))
                    System.out.println(a + " / " + b + " = " + 62);
                
                else 
                    continue;
                
            
        
    

    public static boolean checkNum(int a, int b)
        int A[] = new int[10];
        for (int i=0;i<5;i++)
            A[a%10]++;
            if(A[a%10]>1)
                return false;
            
            a = a/10;
        
        for (int i=0;i<5;i++)
            A[b%10]++;
            if(A[b%10]>1)
                return false;
            
            b = b/10;
        

        for (int i=0;i<10;i++)
            if(A[i]!=1)
                return false;
            
        
        return true;
    

例题2：

public class uva11059 
    public static void main(String[] args) 
//        int n = 3;
//        int A[] = 2,4,-1;
        int n = 5;
        int A[] = 2,5,-1,2,-1;
//        int n = 2;
//        int A[] = -1, 1;
//

        int maxn = -1;
        for (int p1 = 0;p1<=n-1;p1++)
            for (int p2 = 1;p2<=n-p1;p2++)
                int cnt = 1;
                int j = n-p2;
                if (p1==j)
                    cnt = A[p1];
                
                else
                    for (int i=p1;i<=j;i++)
                        cnt *= A[i];
                    
                
                if(cnt>maxn)
                    maxn = cnt;
                
            
        

        System.out.println(maxn>=0?maxn:0);
    

例题3：


代码：

public class uva10976 
    public static void main(String[] args) 
        int k = 2;

        for (int y = k+1;y<=2*k;y++)
            int x1 = k*y/(y-k);
            float x2 = (float) (k*y/(y-k));
            if(Math.abs(x1-x2)<1e-3)
                System.out.println("1/"+k+" = 1/"+x1+" + 1/"+y);