正整數(shù)對（x，y）,x,y都小于等于n,x/y的余數(shù)大于等于k，輸入為n,k,輸出有多少個，比如輸入52,輸出為7,對數(shù)分別為（2，3）(2,4)(2,5)(3,4)(3,5)(4,5)(5,3),代碼如下:importjava.util.Scanner;publicclassMain{publicstaticvoidmain(String[]args){Scannerscanner=newScanner(System.in);intn=scanner.nextInt();intk=scanner.nextInt();System.out.println(searchCount(n,k));}publicstaticintsearchCount(intn,intk){intcount=0;//記錄找到的整數(shù)對個數(shù)inttemp;//思路：固定y，找xfor(inty=k+1;y=k,說明y>k//假設n=a*y+b；在每個長度為y的區(qū)間里只有（y-k）個數(shù)模y余數(shù)>=k。count+=n/y*(y-k);temp=n%y;if(temp>=k){//再考慮余數(shù)b是否>=kcount+=temp-k+1;}}returncount;}}為什么x%y>=k,就能推斷出說明y>k？count+=n/y*(y-k)和count+=temp-k+1這兩句結合為什么就能夠求出最終個數(shù)？