ACM Learning process-bzoj2956 modulus and (number theory)

Description

Ask for ∑∑ ((n mod i) * (M mod j)) where 1<=i<=n,1<=j<=m,i≠j.

　　

Input

The first line is two numbers n,m.

Output

An integer representing the value of the answer mod 19940417

Sample Input

3 4

Sample Output

1

Sample Description
The answer is (3 mod 1) * (4 mod 2) + (3 mod 1) * (4 mod 3) + (3 mod1) * (4 mod 4) + (3 mod 2) * (4 mod 1) + (3 mod 2) * (4 mod 3) + (3 mod 2) * (4mod 4) + (3 mod 3) * (4 mod 1) + (3 mod 3) * (4 mod 2) + (3 mod 3) * (4 mod 4) = 1

Data size and conventions
For data n,m<=10^9 of 100%.

By test instructions:

∑∑ ((n mod i) * (Mmod J)) (I≠j) =∑ (n mod i) *∑ (M mod i)-∑ ((n mod i) * (M mod i)) =∑ (n-[n/i]*i) *∑ (m-[m/i]*i)-∑ ([n /I]+[M/I]) i+[n/i][m/i]*i*i) =∑ (n-[n/i]*i) *∑ (m-[m/i]*i) –n*n*m+∑[n/i]i+∑[m/i]i-∑[n/i][m/i]*i*i (n <= m)

then use [n/i] can be accelerated, but there are a number of intermediate processes that need to be noted,

m/(m/i) the time needed and N Compare size because it may be out of range.

It is also int multiplication may explode, need to turn Long Long , the middle process, don't forget. MOD .

Code:

#include <iostream>#include<cstdio>#include<cstdlib>#include<cmath>#include<cstring>#include<algorithm>#include<Set>#include<map>#include<queue>#include<string>#defineLL Long Long#defineMOD 19940417#defineNsix 3323403using namespacestd;intN, M; LL Cal (intLenintx) {LL ans=0, TMP; intJ;  for(inti =1; I <= Len; ++i) {j= min (len, x/(x/i));//This sentence doesn't have to be min,j to cross .TMP = ((LL) i+j) * (j-i+1)/2%MOD; Ans+ = tmp* (x/i)%MOD; Ans%=MOD; I=J; }    returnans;} inline ll sum (ll x) {returnx* (x+1)%mod* (2*x+1)%mod*nsix%MOD;} LL Cal2 (intXinty) {LL ans=0, TMP, ttmp; intJ;  for(inti =1; I <= x; ++i) {j= Min (x/(x/i), y/(y/i)); //j = min (j, x);TMP = SUM (j)-sum (I-1); TMP= (tmp%mod+mod)%MOD; Ttmp= (LL) x/i) * (y/i)%MOD; Ans+ = tmp*ttmp%MOD; Ans%=MOD; I=J; }    returnans;}voidWork () {if(N >m) Swap (n, m);    LL ans, m2, N2, SNN, SMM, SNM, SS; M2= (LL) m*m%MOD; N2= (LL) n*n%MOD; SMM=Cal (M, m); SNN=Cal (n, N); SNM=Cal (n, m); SS=cal2 (n, m); Ans= M2*n2%mod-m2*snn%mod-n2*smm%mod + snn*smm%MOD; Ans-= m*n2%MOD; Ans+ = m*snn%MOD; Ans+ = n*snm%MOD; Ans-=SS; Ans= (ans%mod+mod)%MOD; printf ("%lld\n", ans);}intMain () {//freopen ("test.in", "R", stdin);     while(SCANF ("%d%d", &n, &m)! =EOF) work (); return 0;}

ACM Learning process-bzoj2956 modulus and (number theory)