牛客网暑期ACM多校训练营（第四场）C（Chiaki Sequence Reloaded）

2024-04-11 06:48:26

题目描述

Chiaki is interested in an infinite sequence a1, a2, a3, ..., which defined as follows:
$a_n=\begin{cases}0 & n = 1 \\ a_{\lfloor \frac{n}{2} \rfloor} + (-1)^{\frac{n(n+1)}{2}} & n \ge 2\end{cases}$
Chiaki would like to know the sum of the first n terms of the sequence, i.e. $\sum\limits_{i=1}^{n}|a_i|$ . As this number may be very large, Chiaki is only interested in its remainder modulo (109 + 7).

输入描述:

There are multiple test cases. The first line of input contains an integer T (1 ≤ T ≤ 105), indicating the number of test cases. For each test case:
The first line contains an integer n (1 ≤ n ≤ 1018).

输出描述:

For each test case, output an integer denoting the answer.

示例1

输入

输出

题意：根据所给公式求绝对值之和

思路：官方题解

直接数位dp记忆化搜索。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
using namespace std;
const int mod=1e9+7;
typedef long long LL;
LL dp[70][3][140];
int dig[70];
int judge (int i,int j)
{
return i==j?1:-1;
}
LL dfs(int pos,int limit,int num,int front)
{
if (pos<0) return abs(70-num);
if (!limit&&dp[pos][front][num]!=-1) return dp[pos][front][num];
int end=limit?dig[pos]:1;
LL res=0;
for (int i=0;i<=end;i++)
{
if (front==2&&i==1) res+=dfs(pos-1,limit&(i==end),num,i);
else if (front==2&&i==0) res+=dfs(pos-1,limit&(i==end),num,2);
else res+=dfs(pos-1,limit&(i==end),num+judge(i,front),i);
res%=mod;
}
if (!limit)
dp[pos][front][num]=res;
return res;
}
int main()
{
memset(dp,-1,sizeof(dp));
LL n;
LL t;
scanf("%lld",&t);
while (t--)
{
scanf("%lld",&n);
int len=0;
while (n) dig[len++]=n&1, n>>=1;
printf("%lld\n",dfs(len-1,1,70,2));
}
return 0;
}

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