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

链接：https://www.nowcoder.com/acm/contest/142/C

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 131072K，其他语言262144K
64bit IO Format: %lld
题目描述
Chiaki is interested in an infinite sequence $a_1, a_2, a_3, ...,$ which defined as follows:

Chiaki would like to know the sum of the first n terms of the sequence, i.e.$\sum_{i=1}^{n}|a_i|$ . As this number may be very large, Chiaki is only interested in its remainder modulo$(10^9 + 7)$.
输入描述:
There are multiple test cases. The first line of input contains an integer $T (1 ≤ T ≤ 10^5)$, indicating the number of test cases. For each test case:
The first line contains an integer $n (1 ≤ n ≤ 10^{18})$.
输出描述:
For each test case, output an integer denoting the answer.
示例1
输入
10
1
2
3
4
5
6
7
8
9
10
输出
0
1
2
2
4
4
6
7
8
11

令n的二进制表示为b(m)b(m-1)…b(0)，且b(m)=1
p(n)是b(i)=b(i+1)的i个数，q(n)是b(i) != b(i+1)的i个数
可以发现|$a_n|=|p(n)+1-q(n)|$
例如$a_2$，$2=(10)_2$，则$p_2=0，q_2=2$，$|a_2|=|0+1-2|=1$
$a_3$，$3=(11)_2$，则$p_3=1，q_2=1$，$|a_3|=|1+1-1|=1$
之后考虑数位dp计算$\sum_{i=1}^{n}|p_i+1-q_i|$

#include<bits/stdc++.h>
using namespace std;
const int MAX=2e5+10;
const int MOD=1e9+7;
typedef long long ll;
inline ll read()
{
    ll f=1,x=0;char ch;
    do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
    do{x=x*10+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');
    return f*x;
}
int p[65];
ll d[65][2][65][65];//d[i][j][p][q]表示在第i位为j的数x中,满足p(x)=p,q(x)=q的数的个数
ll sum[65];
ll cal(ll x)
{
    int n=0;
    while(x)p[++n]=x%2,x/=2;
    ll ans=0;
    int a=0,b=0;
    for(int i=n;i>=1;i--)
    {
        if(i==n){b++;continue;}
        for(int j=0;j<=i;j++)
        {
            if(d[i][0][j][i-j]&&p[i]==1)
            {
                if(p[i+1]==0)ans+=abs(j+a+1-(i-j+b))*d[i][0][j][i-j]%MOD;
                else ans+=abs(j+a-(i-j+b+1))*d[i][0][j][i-j]%MOD;
                if(ans>=MOD)ans%=MOD;
            }
        }
        ans+=sum[i];
        if(ans>=MOD)ans%=MOD;
        if(p[i+1]==p[i])a++;
        else b++;
    }
    ans+=abs(a+1-b);
    if(ans>=MOD)ans%=MOD;
    return ans;
}
int main()
{
    memset(d,0,sizeof d);
    memset(sum,0,sizeof sum);
    for(int i=1;i<=64;i++)
    {
        if(i==1)
        {
            d[i][0][1][0]=1;
            d[i][1][0][1]=1;
            continue;
        }
        for(int j=0;j<=i;j++)
        {
            if(j<i)d[i][1][j][i-j]+=d[i-1][1][j-1][i-j];
            if(i>=j+2)d[i][1][j][i-j]+=d[i-1][0][j+1][i-2-j];
            if(j)d[i][0][j][i-j]+=d[i-1][1][j-1][i-j];
            if(j)d[i][0][j][i-j]+=d[i-1][0][j-1][i-j];
            if(d[i][1][j][i-j]>=MOD)d[i][1][j][i-j]%=MOD;
            if(d[i][0][j][i-j]>=MOD)d[i][0][j][i-j]%=MOD;
        }
        for(int j=0;j<=i;j++)//这里不预处理一下会T。。。
        {
            sum[i]+=abs(j+1-(i-j))*d[i][1][j][i-j]%MOD;
            sum[i]%=MOD;
        }
    }
    int T;
    cin>>T;
    while(T--)
    {
        ll n=read();
        printf("%lld\n",cal(n));
    }
    return 0;
}