HDU 5439 Aggregated Counting

待我称王封你为后i 2022-08-05 10:27 95阅读 0赞

找规律暴力预处理  
之后二分  
1 1  
2 2 3  
3 4 5  
那么5的位置就是1×1\+2×(2\+3)  
然后暴力预处理每一段，复杂度大约是*n*√

// HDU 5439.cpp
    // Created by whn6325689 on 15/9/13.
    // Copyright (c) 2015年 whn6325689. All rights reserved.
    // Mr.Phoebe
    // http://blog.csdn.net/u013007900
    #include <algorithm>
    #include <iostream>
    #include <iomanip>
    #include <cstring>
    #include <climits>
    #include <complex>
    #include <fstream>
    #include <cassert>
    #include <cstdio>
    #include <bitset>
    #include <vector>
    #include <deque>
    #include <queue>
    #include <stack>
    #include <ctime>
    #include <set>
    #include <map>
    #include <cmath>
    #include <functional>
    #include <numeric>
    #pragma comment(linker, "/STACK:1024000000,1024000000")
    
    
    using namespace std;
    
    #define eps 1e-9
    #define PI acos(-1.0)
    #define INF 0x3f3f3f3f
    #define LLINF 1LL<<62
    #define speed std::ios::sync_with_stdio(false);
    
    typedef long long ll;
    typedef unsigned long long ull;
    typedef long double ld;
    typedef pair<ll, ll> pll;
    typedef complex<ld> point;
    typedef pair<int, int> pii;
    typedef pair<pii, int> piii;
    typedef vector<int> vi;
    
    #define CLR(x,y) memset(x,y,sizeof(x))
    #define CPY(x,y) memcpy(x,y,sizeof(x))
    #define clr(a,x,size) memset(a,x,sizeof(a[0])*(size))
    #define cpy(a,x,size) memcpy(a,x,sizeof(a[0])*(size))
    #define debug(a) cout << #a" = " << (a) << endl;
    #define debugarry(a, n) for (int i = 0; i < (n); i++) { cout << #a"[" << i << "] = " << (a)[i] << endl; }
    
    #define mp(x,y) make_pair(x,y)
    #define pb(x) push_back(x)
    #define lowbit(x) (x&(-x))
    
    #define MID(x,y) (x+((y-x)>>1))
    #define ls (idx<<1)
    #define rs (idx<<1|1)
    #define lson ls,l,mid
    #define rson rs,mid+1,r
    
    template<class T>
    inline bool read(T &n)
    {
        T x = 0, tmp = 1;
        char c = getchar();
        while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
        if(c == EOF) return false;
        if(c == '-') c = getchar(), tmp = -1;
        while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();
        n = x*tmp;
        return true;
    }
    template <class T>
    inline void write(T n)
    {
        if(n < 0)
        {
            putchar('-');
            n = -n;
        }
        int len = 0,data[20];
        while(n)
        {
            data[len++] = n%10;
            n /= 10;
        }
        if(!len) data[len++] = 0;
        while(len--) putchar(data[len]+48);
    }
    //-----------------------------------
    
    
    const int MOD=(int)1e9+7;
    const int MAXN=438744;
    int a[MAXN+5];
    int b[MAXN+5];
    int sum[MAXN+5];
    int dp[MAXN+5];
    int pre[MAXN+5];
    int sz,val;
    
    int add(int x,int y)
    {
        x+=y;
        if(x>=MOD)
            x-=MOD;
        return x;
    }
    
    int po(int x,int n)
    {
        int ans=1;
        int temp=x;
        while(n)
        {
            if(n&1)ans=(ll)ans*temp%MOD;
            temp=(ll)temp*temp%MOD;
            n>>=1;
        }
        return ans;
    }
    
    int main()
    {
        val=po(2,MOD-2);
        a[1]=1;a[2]=2;a[3]=2;
        int sz=3;
        for(int i=3;;i++)
        {
            int cnt=a[i];
            while(cnt)
            {
                a[++sz]=i;
                cnt--;
                if(sz>=MAXN)break;
            }
            if(sz>=MAXN)    break;
        }
        for(int i=1; i<=MAXN; i++)
        {
            sum[i]=sum[i-1]+a[i];
            if(sum[i]>=INF) break;
        }
    
        for(int i=1; i<=MAXN; i++)
        {
            dp[i]=add(dp[i-1],(ll)( add(add(add(pre[i-1],1),pre[i-1]),a[i]) )*a[i]%MOD*i%MOD*val%MOD);
            pre[i]=add(pre[i-1],a[i]);
        }
    
        int T;
        read(T);
        while(T--)
        {
            int n;
            read(n);
            int l=1,r=MAXN;
            int pos=0;
            while(l<=r)
            {
                int mid=MID(l,r);
                if(n<=sum[mid])
                {
                    pos=mid;
                    r=mid-1;
                }
                else l=mid+1;
            }
            int x=n-sum[pos-1];
            int ans=add(dp[pos-1],(ll)( add(add(add(pre[pos-1],1),pre[pos-1]),x) )*x%MOD*pos%MOD*val%MOD);
            printf("%d\n",ans);
        }
        return 0;
    }