求解代码
publicstaticvoidmain(String[]args)throwsIOException{BufferedReaderbr=newBufferedReader(newInputStreamReader(System.in));StreamTokenizerin=newStreamTokenizer(br);PrintWriterout=newPrintWriter(newOutputStreamWriter(System.out));in.nextToken();intn=(int)in.nval;// 创建HashMap：键=差值diff（Integer），值=该diff出现的次数（Integer）Map<Integer,Integer>hMap=newHashMap<>();for(inti=0;i<n;i++){in.nextToken();inta=(int)in.nval;// diff = 前整数 - 它的位置索引intdiff=a-i;hMap.put(diff,hMap.getOrDefault(diff,0)+1);}longtotal=0;// 遍历Map中所有diff的出现次数for(intcount:hMap.values()){// 只有出现次数>1的diff，才能组成数对（至少2个元素才能配对）if(count>1){// 相同 diff 的数，两两之间都满足题目要求的条件，// 所以统计每个 diff 的出现次数，再算组合数就是答案。total+=(long)count*(count-1)/2;}}out.println(total);out.flush();out.close();br.close();}小贴士
a j − a i = j − i a_j -a_i = j-iaj−ai=j−i➡️a j − j = a i − i a_j - j= a_i-iaj−j=ai−i
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