算法基础——1.2枚举与剪枝

来源：互联网发布：夜猫子软件编辑：程序博客网时间：2024/04/29 14:54

例一

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/* 
  要找8元钱。 
  有零钞：5元，2元，1元，5角。 
  求所有找零方案 
*/  
  
public class T1  
{  
    public static void main(String[] args)  
    {  
        int SUM = 80;  
          
        for(int a=0; a <= SUM/50; a++){  
            for(int b=0; b <= SUM/20; b++){  
                for(int c=0; c <= SUM/10; c++){  
                    for(int d=0; d <= SUM/5; d++){  
                        if(a*50 + b*20 + c*10 + d*5==SUM)  
                            System.out.println(a+","+b+","+c+","+d);  
                    }                 
                }  
            }  
        }  
    }  
}  

例二:

[java] view plaincopy
/* 
 n位数平方的尾数还是自己 
*/  
  
public class T2  
{  
    public static void main(String[] args)  
    {  
        // 仔细观察  
        // 这样的数字的尾数有什么特征？ 0 1 5 6  
        // 进一步，其末尾2位数必须是 25 76  
        // 更进一步，其末尾3位数必须是？。。。。  
          
        for(int i=100; i<1000; i++){  
            int n = i*i;  //平方后  
            int m = i%10;  
            if(m!=0 && m!=1 && m!= 5 && m!= 6) continue;              
            if(n%1000==i) System.out.println(i + "," + n);  
        }  
    }  
}  

例三：

方法一

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/*观察算式 
观察下面的算式： 
△△△ * △△ = △△△△ 
某3位数乘以2位数，结果为4位数 
要求：在9个△所代表的数字中，1~9的数字恰好每个出现1次。*/  
public class HomeWork2 {  
  
    public static void main(String[] args) {  
        long startTime=System.nanoTime();//获取开始时间  
          
        int a , b, c;//a*b=c    a = i1 * 100 + i2 * 10 + i3;  b = i4 * 10 + i5;  
        for(int i1 = 1; i1 < 10; ++i1)  
            for(int i2 = 1; i2 < 10 ; ++i2){  
                if(i2 == i1)  
                    continue;  
                for(int i3 = 1; i3 < 10 ; ++i3)  
                {  
                    if(i3 == i1 || i3 == i2)  
                        continue;  
                      
                    a = i1 * 100 + i2 * 10 + i3;  
                  
                    for(int i4 = 1; i4 < 10 ; ++i4){  
                        if(i4 == i1 || i4 == i2 || i4 == i3)  
                            continue;  
                        for(int i5 = 1; i5 < 10 ; ++i5)  
                        {  
                            if(i5 == i1 || i5 == i2 || i5 == i3 || i5 == i4)  
                                continue;  
                            b = i4 * 10 + i5;  
                            c = a * b;  
                              
                            if(c <= 9876 && c >= 1234 )  
                            {  
                                int c1 = c/1000;//c的千位  
                                int c2 = (c/100)%10;//c的百位  
                                int c3 = (c/10)%10;//c的十位  
                                int c4 = c%10;//c的个位  
                                //c不满足条件  
                                if(c1 == 0 || c2 == 0 || c3 == 0 || c4 == 0|| c1 == c2 || c1 == c3 || c1 == c4 || c2 == c3 || c2 == c4 || c3 == c4  || c1 == i1 || c1 == i2 || c1 == i3 || c1 == i4 || c1 == i5 || c2 == i1 || c2 == i2 || c2 == i3 || c2 == i4 || c2 == i5|| c3 == i1 || c3 == i2 || c3 == i3 || c3 == i4 || c3 == i5 || c4 == i1 || c4 == i2 || c4 == i3 || c4 == i4 || c4 == i5)  
                                    continue;  
                                else System.out.println(a+" * "+b+" = "+c);  
                            }                     
                        }  
                    }  
                }  
            }  
        long endTime=System.nanoTime(); //获取结束时间  
        System.out.println("程序运行时间： "+(endTime-startTime)+"ns");  
  
  
    }  
  
}  

方法二

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public class HomeWork1 {  
  
        /** 
         * 观察下面的算式：△△△ * △△ = △△△△ 某3位数乘以2位数，结果为4位数。 
         * 要求：在9个△所代表的数字中，1~9的数字恰好每个出现1次。 
         */  
        public static void main(String[] args) {  
            long startTime=System.nanoTime();  
            int num[]= new int[10]; //只使用num[1]~num[9],做1~9数字的哈希映射表。  
  
            for(int i=12; i<=98; i++){  
                //两位数最小从12开始，最大98  
                if(i%10 == 0 || i%10 == i/10 )  
                    continue;//末位为零，或两位相等，剪掉  
                initArray(num, 0);      //初始化哈希映射表，表元素num[1]~num[9]全为零  
                num[i%10]=1;//将两位数的两个数字在哈希映射表中做标志，为1  
                num[i/10]=1;  
                for(int j= 1;j<=9; j++){  
                    //遍历哈希映射表num[],找出所有满足条件的三位数，每次从最小的三位数找起。  
                    if(num[j]==0){//j为三位数的百位  
                        num[j]=2;//将满足的j标记为2  
                        for(int k=1;k<=9;k++){//k为三位数的十位数，  
                            if(num[k]==0){  
                                num[k]=3;//将满足的k标记为3  
                                for(int n=1;n<=9;n++){//n为三位数的各位数  
                                    if(num[n]==0){  
                                        num[n]=4;//将满足的n，标记为4  
                                        int thirdnum = j*100+ k*10+ n;//符合条件的三位数  
                                        int fact=i*thirdnum;  
                                        int fact1=fact;  
                                        if(fact>9876){//两位数乘以三位数若大于最大的四位数，则剪掉  
                                            num[n]=0;//复原已标记的个位数  
                                            break;  
                                        }  
                                        boolean flag=true;//检查得到的四位数满不满足条件  
                                        for(int m=1000;m>=1;m/=10){  
                                            if(num[fact1/m]==0){  
                                                num[fact1/m]=-1;//在哈希映射表中查找是否是剩余的数字，若是，标记为-1  
                                                fact1=fact1-(fact1/m)*m;  
                                            }else{  
                                                flag=false;  
                                                break;  
                                            }  
                                        }  
                                        if(flag == true){//满足则输出  
                                            System.out.println("满足条件的算式是：");  
                                            System.out.println(i + " * " + thirdnum + " = " + fact);  
                                        }  
                                        num[n]=0;//复原已标记的个位数  
                                        initArray(num, -1);//复原已标记的-1  
                                    }  
                                }  
                                  
                                num[k]=0;//复原已标记的十位数  
                            }  
                        }  
                        num[j]=0;//复原已标记的百位数  
                    }  
                }  
            }  
            long endTime=System.nanoTime(); //获取结束时间  
            System.out.println("程序运行时间： "+(endTime-startTime)+"ns");  
        }  
        public static void initArray(int num[], int flag){  
            //初始化映射数组,当flag=0，清零num[1]~num[9]数组元素,flag=-1,1,2,3,4，则清零值等于flag的数组元素。  
            switch(flag){  
            case -1:  
            case 1:  
            case 2:  
            case 3:  
            case 4: for(int i=0;i<num.length;i++){     
                        if(num[i] == flag){  
                            num[i]=0;  
                        }  
                    }  
                    num[0]=-2;//num[0]不用，标记为-2  
                    break;  
            default:for(int i=0;i<num.length;i++){     
                        num[i]=0;  
                    }  
                    num[0]=-2;//num[0]不用，标记为-2  
            }  
        }  
}  

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