HDU5439 Aggregated Counting （找规律+预处理+二分）

Aggregated Counting

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 666 Accepted Submission(s): 302

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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≤10

9

)


representing a query.

[align=left]Output[/align]
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

.

[align=left]Sample Input[/align]

3
3
10
100000

[align=left]Sample Output[/align]

11
217
507231491

[align=left]Source[/align]
2015 ACM/ICPC Asia Regional Changchun Online

题意：f(n)表示序列中n的最后的位置，求f（f(n)）。
分析：可以令n=1+2+2+3+3+......+ i 这个序列的长度为p

那么a
=1*1+2*2+3*2+...... + p*i

那么不难发现a[a
] = 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 是 第p-1项的pre + b[p-1]
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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 1e9;
const int MOD = 1e9+7;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define lson (i<<1)
#define rson ((i<<1)|1)
#define MAXN 438744

int a[MAXN+5],b[MAXN+5];
int sum[MAXN+5],dp[MAXN+5],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;
}