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
which defined as follows:
Chiaki would like to know the sum of the first n terms of the sequence, i.e.
. As this number may be very large, Chiaki is only interested in its remainder modulo
.
输入描述:
There are multiple test cases. The first line of input contains an integer
, indicating the number of test cases. For each test case:
The first line contains an integer
.
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|
例如
,
,则
,
,
,则
,
之后考虑数位dp计算
#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;
}