"Solving" A-b

"Problem description"
It's a painful thing to have a quiz!
Topic see more also have aesthetic fatigue, so I abandoned everyone familiar a+b problem, switch to a-B ha
Ha!
Well, here's the question: give a bunch of numbers and a number C, which requires that all a-b=c be counted.
Number. (number pairs of numbers in different positions are counted in different pairs)
Input
The first line consists of 2 non-negative integers N and C, separated by a space in the middle.
The second line has N integers, separated by a space, as the number of strings required to be processed.
Output
Outputs a row that represents the number of all the pairs of a-b=c that are included in the number of strings.
"Input and Output sample"

Dec.in Dec.out
4 1 3
1 1 2 3
First, such as 2,3,4,6,7, to make the difference of 1.
You find that 3-2 equals 1, then 4-2 must be greater than 1, so to reduce to 1, only 2 behind that number is possible.
So if 4-3 equals 1, then 6 must start from the 3 behind that number, so the problem is very simple.
Of course, if there is repetition, you still need to record it.

1#include <algorithm>2#include <iostream>3#include <fstream>4#include <cstdlib>5 /*Run this program using the console Pauser or add your own getch, System ("pause") or input loop*/6 using namespacestd;7 8Ifstream Fin ("dec.in");9Ofstream Fout ("Dec.out");Ten  One intcnt_shu=0; A Long LongCha=0; - Long Longshu[200005]={0}; - intcnt_cs[200005]={0}; the Long Longsze_jianshu=0; - Long Longcs=0; -  - BOOLQqsort (Long LongALong Longb); + voidXunintSze_jian); -  + BOOLQqsort (Long LongALong Longb) { A  if(A&LT;B)return 1; at  return 0;  - } -  -  - voidXunintSze_jian) { -  if(Sze_jianshu==sze_jian) sze_jianshu++;  in   for(intx=sze_jianshu;x<=cnt_shu;x++){ -   if(shu[x]-shu[sze_jian]==cha) { tocs+=cnt_cs[sze_jian]*Cnt_cs[x]; +sze_jianshu=x;  -   }       the   if(Shu[x]-shu[sze_jian]<cha) sze_jianshu=x; *   Else  Break; $  }    Panax Notoginseng  return; - }  the  + intMainintargcChar*argv[]) { AFin>>cnt_shu>>cha; the  if(cha<0) cha=0-cha; +   for(intx=1; x<=cnt_shu;x++) fin>>Shu[x]; -Sort (shu+1, shu+1+cnt_shu,qqsort); $   $  intDa=1, gs=1, num=shu[1]; -   for(intx=2; x<=cnt_shu+1; x + +){ -   if(shu[x]==num&&x!=cnt_shu+1){ thegs++; -    Continue;Wuyi   }      theshu[da]=num;  -num=Shu[x]; Wucnt_cs[da++]=GS; -gs=1; About  } $   -  if(cha==0){ -   Long Longcs1=0; -    for(intx=1; x<da;x++){ Acs1+= (cnt_cs[x]* (cnt_cs[x]-1))/2; +   }     thefout<<CS1; -   return 0;  $  } the   thesze_jianshu=1;  the   for(intx=1; x<da;x++){ the Xun (x);  -  } infout<<CS; the  return 0; the}

"Solving" A-B