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* integer arr be divided into pa

* integer arr be divided into pa

作者: 程序员编程助手 | 来源:发表于2025-05-16 17:29 被阅读0次

/**

* integer arr be divided into pairs

* return pairs number depend on

* parameter k : key is a factor of pairs sum multiply element

* constraints

* 2<=n<=100 arr size

* 1<=k<=100 multiply element factor

* 1<=ar[i]<=100 arr element value

*

* @param n

* @param k

* @param ar

* @return

*/

public static int divisibleSumPairs(int n,int k,Listar) {

if (n <2 ||n >100) {

return 0;

}

if (k <1 ||k >100) {

return 0;

}

if (ar ==null) {

return 0;

}

if (ar.isEmpty()) {

return 0;

}

Random random =new Random();

int count =0;

ArrayListstringArrayList2 =new ArrayList<>();

while (true) {

int arrIndex =random.nextInt(ar.size());

int arrIndexNextSeeds =random.nextInt(ar.size());

if (!(arrIndexNextSeeds ==arrIndex)) {

Integer i =ar.get(arrIndex);

Integer i1 =ar.get(arrIndexNextSeeds);

stringArrayList2.add(arrIndex+"-"+arrIndexNextSeeds+"-"+i+"-"+i1);

}

count++;

if (count >1000000) {

break;

}

}

HashSetstringHashSet =new HashSet<>();

for (String s :stringArrayList2) {

stringHashSet.add(s);

}

ArrayListstringArrayList1 =new ArrayList<>();

for (String e :stringHashSet) {

String[]split =e.split("-");

String s2 =split[2];

String s3 =split[3];

int i =Integer.valueOf(s2) +Integer.valueOf(s3);

int i1 =i %k;

if (i1 ==0) {

stringArrayList1.add(e);

}

}

return stringArrayList1.size()/2;

}

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