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  • 2019牛客多校九 I. KM and M (类欧几里得)

    大意: 给定$N,M$, 求$sumlimits_{K=1}^N ext{(KM)&M}$

    考虑第$i$位的贡献, 显然为$lfloorfrac{KM}{2^i} floor$为奇数的个数再乘上$2^i$

    也就等于$2^i(sum lfloorfrac{KM}{2^i} floor-2sum lfloorfrac{KM}{2^{i+1}} floor)$, 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 '
    '
    #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
    ", ans);
    }
    
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  • 原文地址:https://www.cnblogs.com/uid001/p/11416582.html
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