Han soldiers optimization problem (Java algorithm)

 

1, the problem description

　　① Huai'an folklore with a story - " Han Xin Troops", followed by a phrase "Han soldiers, the more the better ."

Han fight with 1,500 soldiers, killed four to five people standing in a row 3, extra 2; 5 stand in a row, the extra 4; 7 stand in a row, six more people. Han soon to say the number: 104.

　　② a thousand years ago, " Sun Tzu Suan Jing ", there is such an arithmetic problem: "Today there was an unknown number, the number of remaining two hundred thirty-three, fifty-five number of remaining three numbers of the remaining two hundred seventy-seven, geometric asked was "in accordance with today's words:? a number is divided by 32, is divided by 53, divided by more than 72, find the number. This problem was also known as "Han soldiers." It forms a class of problems, that is, elementary number theory solutions of the congruence.

　　③ problem: a number is divided by 32, is divided by 53, is divided by 72, to find the minimum number of matches.

2, concluding ideas

1 idea: [violent solving method - low traversal, efficiency]

From 0 up to a large value is calculated to be judged

    public static int dealSingle1(int index0,int index1,int index2){//暴力求解
        boolean a,b,c;
        for(int i=0;i<Integer.MAX_VALUE;i++){
            a=(i%3==index0);
            b=(i%5==index1);
            c=(i%7==index2);
            if(a&&b&&c){
                return i;
            }
        }
        return -1;
    }

 

2 thoughts: [the use of mathematical knowledge by calculating a further optimization, it should be the fastest]

　　Listed first number is divided by 32: 2,5,8,11,14,17,20,23,26 ......

　　And then lists the number is divided by 53: 3,8,13,18,23,28 ......

　　Numbers in these two columns, the number of common First there is the 8.3 and 5 the least common multiple of 15. Two conditions are combined into a is an integer of 8 + 15 ×, which lists the number string is 8,23,38, ......, and then divided by the number of lists of 2,9,16,23,30 I ...... 72 to come in line with the minimum number of title conditions is 23.

　　In fact, we have already the title three conditions are combined into one: in addition to more than 105 by 23.

    public static int dealSingle2(int index0,int index1,int index2) {//2,数学方法求解
        long startTime =  System.currentTimeMillis();
        int num=0,num1=0;
        for(int i=0;i<Integer.MAX_VALUE;i++){
            num=3*i+index0;
            if(num%5==index1){
                break;
            }
        }
        for(int i=0;i<Integer.MAX_VALUE/15;i++){
            num1+ 15 * NUM = I;
             IF (num1% ==. 7 index2) {
                 Long endTime =   System.currentTimeMillis (); 
                System.out.println ( "Method 2: mathematical optimization, Processed:" + (endTime- the startTime)) ;
                 return num1; 
            } 
        } 
        return -1 ; 
    }

3 ideas: [dominated by the large numbers numerical computation, very fast]

    public static int dealSingle3(int index0,int index1,int index2){//处理单个数据
        long startTime =  System.currentTimeMillis();
        int num=-1;
        boolean a,b;
        for(int i=0;i<Integer.MAX_VALUE/7;i++){
            num=i*7+index2;
            a=(num%3==index0);
            b=(num%5==index1);
            if(a&&b){
                long endTime =  System.currentTimeMillis (); 
                System.out.println ( "Method 3: mathematical optimization, Processed:" + (endTime- the startTime));
                 return NUM; 
            } 
        } 
        return NUM; 
    }

 3, the problem extended

　　All calculated results of [3,5,7 Han soldiers all prime numbers, then 0 ~ (3 * 5 * 7-1) all data within this range is a situation in which]

    public  static  int [] [] [] dealAll () { // each calculation only once, until all data 
        int [] [] [] = ARR new new  int [. 3] [. 5] [. 7 ];
         for ( int I = 0; I <(. 5 * *. 7. 3); I ++ ) { 
            ARR [I %. 3] [I. 5%] [% I. 7] = I; 
        } 
        return ARR; 
    }