NYOJ 508 remainder summation (number theory problem)

NYOJ 508 remainder summation (number theory problem)

Description

 

Give you two numbers n, k, the requested value.

Input two numbers n, k (1 <= n, k <= 10 ^ 9) in each row. Output and, each result occupies one row. Sample Input

5 45 3

Sample output

57

Question Analysis:

This question cannot be directly violent. It has a large value and can only use the sqrt (n) algorithm to calculate the remainder sum of n % I, I = 1 ~ N; Note: n % I = n/I * I;

Therefore, we can simulate ~ Sqrt (n); sum + = n % I; for j = n/I corresponding to I, you can know (n/I ~ N/I + 1) = d; for the number of groups

S1 = (n/I-n/(I + 1) * n)-d * (n/(I + 1) + 1 + ...... + n/I); sum + = s1; after the loop ends, the remainder sum of n % I is obtained. We use ModSum (n ).

Now we calculate the remainder sum of k % I, I = 1 ~ N; analyze k and n,

I. k K % n = k

II. k = n res = ModSum (k)

3. k> n res = ModSum (k); {for (int I = n + 1; I <= n; I ++)

Res-= k % I; // subtract the value of multiple computations.

}

 

AC code:

 

 /** *1、ModSum1()：O(n) *2、ModSum2()：O(sqrt(n)) *1   2   3   4 *16  8   5   4 *  1   2   3 */#include
   
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               #define LL long long#define MOD 1000000007using namespace std;LL ModSum1(LL n){ LL t=(n+1)/2-1; LL sum=t*(t+1)/2; for(LL i=1;i<=n/2;i++){ sum+=n-(n/i*i); } return sum;}LL ModSum2(LL n,LL k){ LL sum=0,n1=n; n=k; LL tem=sqrt(n); for(LL i=1;i<=tem;i++){ if(n/i==i){ sum+=n-n/i*i; continue; } sum+=n-n/i*i; sum=sum; //cout<
               
                >n>>k){ //LL sum1=ModSum1(n); LL sum2; sum2=ModSum2(n,k); cout<