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);
}