Xiangtan OJ1203/Invitational A title number theory +java Eyelid plate

Begging n%1+n%2+n%3+n%4+.........n%n=,n<=10^12 secondary.

Blindly looking for the law of the beginning. Not found. Later, after the senior coach, talent insipid, so, now the only clear.

First on the map:

For this problem, in the interval (root n,n), because n%i=n-i*x (here X is from 1 to the square root n, each k corresponding n/(x+1) ~n/x interval. Because it is arithmetic progression (or descending). Sum directly with the formula).

Ah (root n,n) interval is to be cut to obtain. Divided into square root n times.

Import Java.io.*;import Java.util.scanner;import java.math.biginteger;public class main{public static void Main (String    [] args) {int T;    Scanner in=new Scanner (system.in);    T=in.nextint ();                                              for (int ii=1;ii<=t;ii++) {long n;           BigInteger sum=new BigInteger ("0");           N=in.nextlong ();        Long lasta=2*n;         for (long i=1;i*i<n;i++) {Sum=sum.add (biginteger.valueof (n%i));              if ((i+1) >=lasta) {break;}              Boundary inference long b=n/i;              Long a=n/(i+1) +1;         BigInteger temp1=biginteger.valueof (n);                 Convert a Long data to BigInteger temp1=temp1.multiply (biginteger.valueof (b-a+1));              BigInteger temp2=biginteger.valueof (b+a);    Temp2=temp2.multiply (biginteger.valueof (i));                 * Temp2=temp2.multiply (biginteger.valueof (b-a+1));   Temp2=temp2.divide (biginteger.valueof (2));       Dividing       Temp1=temp1.subtract (TEMP2);                             -Sum=sum.add (TEMP1);         + lasta=a;         } System.out.println ("Case" +ii+ ":" +sum); The Java println itself takes the initiative to bring a newline.    Attention! }}}

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Xiangtan OJ1203/Invitational A title number theory +java Eyelid plate