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  • Codeforces 1151C Problem for Nazar

    大意: 给定一个无限长的序列, 求区间和.

    这题比赛的时候改半天没改出来, 以后做这种模拟应该把思路理清再写.

    就是利用分块的思想, 预处理出每一块的和, 再询问时用中间整块的和加上左右部分块的和.

    #include <iostream>
    #include <algorithm>
    #include <cstdio>
    #include <math.h>
    #include <set>
    #include <map>
    #include <queue>
    #include <string>
    #include <string.h>
    #include <bitset>
    #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, INF = 0x3f3f3f3f;
    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
    
    
    
    #ifdef ONLINE_JUDGE
    const int N = 1e6+10;
    #else
    const int N = 111;
    #endif
    
    const int mx = 62;
    int cnt;
    ll a[N], st[N], sum[N], inv2 = (1+P)/2;
    
    int blo(ll x) {
    	return lower_bound(a+1,a+1+mx,x)-a;
    }
    ll L(ll x) {
    	return a[blo(x)-1]+1;
    }
    ll R(ll x) {
    	return a[blo(x)];
    }
    ll num(ll x) {
    	ll n = x-L(x)+1;
    	return (st[blo(x)]+(n-1)*2)%P;
    }
    void init() {
    	REP(i,1,mx) a[i] = (1ll<<i)-1;
    	st[1]=1,st[2]=2;
    	REP(i,3,mx) st[i] = (st[i-2]+(1ll<<i-2))%P;
    	sum[1]=1,sum[2]=7;
    	REP(i,3,mx) {
    		ll n = (1ll<<i-1)%P;
    		sum[i] = (sum[i-1]+n*st[i]%P+n*(n-1)%P)%P;
    	}
    }
    
    ll solve(ll l, ll r) {
    	ll ans = 0;
    	if (blo(l)==blo(r)) { 
    		ans = (num(r)+num(l))%P*(r%P-l%P+1)%P*inv2%P;
    	}
    	else ans=(solve(l,R(l))+solve(L(r),r)+sum[blo(r)-1]-sum[blo(l)])%P;
    	return ans;
    }
    
    int main() {
    	init();
    	ll l, r;
    	cin>>l>>r;
    	ll ans = solve(l,r);
    	if (ans<0) ans+=P;
    	cout<<ans<<endl;
    }
    
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  • 原文地址:https://www.cnblogs.com/uid001/p/10733842.html
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