codevs2606 Approximate and problem

Title Description Description

Smart has recently been addicted to the study of approximations.

For a number x, the function f (x) represents the and of all the approximations of X. For example: F (6) =1+2+3+6=12. For a x,smart you can quickly figure out F (X). The problem now is that, given two positive integers, x, y (x<y), Smart wants to figure out the value of f (X) +f (x+1) +......+f (Y) as soon as possible, can you help smart figure out the value?

Enter a description input Description

The input file is only one line, two positive integers X and y (x<y), indicating that f (X) +f (x+1) +......+f (Y) needs to be computed.

outputs description output Description

The output is only one row and is the value of f (X) +f (x+1) +......+f (Y) .

sample input to sample

2 4

Sample output Sample outputs

14

data size & Hint

There is 1≤x

There is 1≤x

There is 1≤x

Positive solutions: chunking

Problem Solving Report:

　　is said to be a universal group problem, I actually thought for so long, no.

The sum of the approximate sum of the interval [l,r], and the direct to the endpoint subtraction.  Then consider the answer must be ans=∑[n/i]*i (1<=i<=n); But we don't have to. For once all I, can be [n/i] equal interval processing (block processing), directly to this interval sum is OK.

1 //It's made by jump~2#include <iostream>3#include <cstdlib>4#include <cstring>5#include <cstdio>6#include <cmath>7#include <algorithm>8#include <ctime>9#include <vector>Ten#include <queue> One#include <map> A#include <Set> - using namespacestd; -typedefLong LongLL; the //ans=∑[n/i]*i (1<=i<=n); by [n/i] chunking -  -InlineintGetint () - { +        intw=0, q=0;CharC=GetChar (); -         while((c<'0'|| C>'9') && c!='-') C=getchar ();if(c=='-') q=1, c=GetChar (); +         while(c>='0'&& c<='9') w=w*Ten+c-'0', C=getchar ();returnQ? -w:w; A } at  - inline ll solve (ll N) { -     if(n==0|| n==1)returnN; -LL left=1, right; LL ans=0; -      while(left<=N) { -right=n/(N/left);//determine the right endpoint for the same value [n/i] inans+= (n/left) * (left+right) * (right-left+1)/2; -left=right+1; to     } +     returnans; - } the  *InlinevoidWork () { $LL x, y; X=getint (); y=getint ();Panax Notoginsengprintf"%lld", Solve (y)-solve (X-1)); - } the  + intMain () A { the Work (); +   return 0; -}

codevs2606 Approximate and problem