2019牛客多校九 I. KM and M (类欧几里得)

大意: 给定$N,M$, 求$\sum\limits_{K=1}^N \text{(KM)&M}$

考虑第$i$位的贡献, 显然为$\lfloor\frac{KM}{2^i}\rfloor$为奇数的个数再乘上$2^i$

也就等于$2^i(\sum \lfloor\frac{KM}{2^i}\rfloor-2\sum \lfloor\frac{KM}{2^{i+1}}\rfloor)$, 可以用类欧求出

#include <iostream>

#include <sstream>

#include <algorithm>

#include <cstdio>

#include <cmath>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <cstring>

#include <bitset>

#include <functional>

#include <random>

#define REP(i,a,n) for(int i=a;i<=n;++i)

#define PER(i,a,n) for(int i=n;i>=a;--i)

#define hr putchar(10)

#define pb push_back

#define lc (o<<1)

#define rc (lc|1)

#define mid ((l+r)>>1)

#define ls lc,l,mid

#define rs rc,mid+1,r

#define x first

#define y second

#define io std::ios::sync_with_stdio(false)

#define endl '\n'

#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int P = 1e9+7, inv2 = (P+1)/2;

ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}

ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}

inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}

//head

int n, m;

int solve(ll a, ll b, ll c, ll n) {

    if (!a) return (n+1)*(b/c)%P;

    if (a>=c||b>=c) {

        int t = solve(a%c,b%c,c,n); n %= P;

        return (t+(a/c)%P*n%P*(n+1)%P*inv2+(b/c)%P*(n+1))%P;

    }

    ll m = ((__int128)a*n+b)/c;

    return ((n%P)*(m%P)-solve(c,c-b-1,a,m-1))%P;

}

int main() {

    ll n, m;

    cin>>n>>m;

    int ans = 0;

    REP(i,0,60) if (m>>i&1) {

        int A = solve(m,0,1ll<<i,n);

        int B = solve(m,0,1ll<<i+1,n);

		int C = (1ll<<i)%P;

        ans = (ans+(A-2ll*B)*C)%P;

    }

    if (ans<0) ans += P;

    printf("%d\n", ans);

}