Algorithm Note _152: algorithm to improve fu Granny Cross Street (Java)

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1 Problem Description

2 Solutions

  1 problem description

A total of 5 red scarves, numbered A, B, C, D, E, the grandmother was one of them to help cross the road.

Five red scarves speak each other:

A: Neither I nor e have helped granny

B: Granny is a C and E, one of them helped me through the street.

C: Granny's been helped by me and D.

D:b and C are not helping granny across the street.

E: I didn't help grandma.


It is known that there are five red scarves and only 2 people are telling the truth, who helped the old lady through the street?

If there are multiple answers, output in one line, separated by a space between the numbers.


For example
A B C D E (which is obviously not the correct answer)

2 Solutions

The specific code is as follows:

Importjava.util.ArrayList;Importjava.util.Collections; Public classMain { Public Staticarraylist<integer> result =NewArraylist<integer>();  Public voidGetResult (intIintj) {//Ans[i] = 0 indicates the initial state, ans[i] = 1 indicates that there must be no help//Ans[i] = 1 means there must be help, ans[i] = 2 indicates that it is possible to help        int[] ans =New int[5]; //a says        if(i = = 0 | | j = 0) {ans[0] = 1; ans[4] = 1; } Else{ans[0] = 2; ans[4] = 2; }        //b says        if(i = = 1 | | j = 1) {ans[2] = 2; if(Ans[4] = = 2) {ans[4] = 1; } Else if(Ans[4] = =-1) ans[2] = 1; } Else{ans[2] = 1; if(Ans[4] = = 2) {ans[4] = 1; ans[0] = 1; }        }        //c says        if(i = = 2 | | j = 2) {ans[3] = 2; if(Ans[2] = = 2) ans[2] = 1; Else if(ans[2] = =-1) ans[3] = 1; } Else {            if(Ans[2] = = 1)                return; Elseans[2] = 1; ans[3] = 1; }        //D says        if(i = = 3 | | j = 3) {ans[1] = 1; if(Ans[2] = = 1)                return; Else if(i = = 2 | | j = 2) {ans[2] = 1; ans[3] = 1; } Else if(i = = 1 | | j = 1) {ans[2] = 1; ans[4] = 1; }        } Else{ans[1] = 2; if(Ans[2] = = 2) ans[2] = 1; Else if(ans[2] = =-1) ans[1] = 1; }        //e says        if(i = = 4 | | j = 4) {            if(Ans[4] = = 1)                return; Elseans[4] = 1; } Else {            if(Ans[4] = =-1)                return; Elseans[4] = 1; }        intCount = 0, temp = 0;  for(intt = 0;t < 5;t++) {            if(Ans[t] = = 1) {Count++; Temp=T; }        }        if(Count = = 1) {            if(!result.contains (temp)) Result.add (temp); }        return; }         Public Static voidMain (string[] args) {main test=NewMain ();  for(inti = 0;i < 5;i++) {             for(intj = i + 1;j < 5;j++) {Test.getresult (i, j);        }} collections.sort (Result);  for(inti = 0;i < Result.size (); i++) {            Chartemp = (Char) (' A ' +Result.get (i)); System.out.print (Temp+" "); }    }}

Algorithm Note _152: algorithm to improve fu Granny Cross Street (Java)