2015长春网络赛1003（hdu5439）推公式

Aggregated Counting

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 50    Accepted Submission(s): 19

Problem Description

Aggregated Counting Meetup (ACM) is a regular event hosted by Intercontinental Crazily Passionate Counters (ICPC). The ICPC people recently proposed an interesting sequence at ACM2016 and encountered a problem needed to be solved.

The sequence is generated by the following scheme.
1. First, write down 1, 2 on a paper.
2. The 2nd number is 2, write down 2 2’s (including the one originally on the paper). The paper thus has 1, 2, 2 written on it.
3. The 3rd number is 2, write down 2 3’s. 1, 2, 2, 3, 3 is now shown on the paper.
4. The 4th number is 3, write down 3 4’s. 1, 2, 2, 3, 3, 4, 4, 4 is now shown on the paper.
5. The procedure continues indefinitely as you can imagine. 1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, . . . .

The ICPC is widely renowned for its counting ability. At ACM2016, they came up with all sorts of intriguing problems in regard to this sequence, and here is one: Given a positive number  n, First of all, find out the position of the last  n that appeared in the sequence. For instance, the position of the last 3 is 5, the position of the last 4 is 8. After obtaining the position, do the same again: Find out the position of the last (position number). For instance, the position of the last 3 is 5, and the position of the last 5 is 11. ICPC would like you to help them tackle such problems efficiently.

 

Input

The first line contains a positive integer  T,T≤2000, indicating the number of queries to follow. Each of the following  T lines contain a positive number  n(n≤109) representing a query.

 

Output

Output the last position of the last position of each query  n. In case the answer is greater than  1000000006, please modulo the answer with  1000000007.

 

Sample Input

3 3 10 100000

 

Sample Output

11 217 507231491

 

题解：

 可以令n=1+2+2+3+3+......+ i    这个序列的长度为p

那么a[n]=1*1+2*2+3*2+...... + p*i 

那么不难发现a[a[n]] = 1*1 + (2+3)*2 + (4+5)*3 + (6+7+8)*4 + ... + (pre+1 + pre+2 + ... + pre+b[p] ) * p

b[p]为p在原序列中出现的次数

pre，b[p]这些值都可以预处理算出   每次询问可以O（1）算出答案

pre 是 第p-1项的pre + b[p-1]

                比赛没时间想了哎，赛后一推就出来了，果然还是太弱- -！

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

typedef long long ll;
const int MOD = (int)1e9+7;
const int INF = (int)1e9;
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;scanf("%d",&t);
    while(t--){
        int n;scanf("%d",&n);
        int l=1,r=MAXN;
        int pos;
        while(l<=r){
            int mid=l+r>>1;
            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;
}