Yousef loves playing with functions in his free time. Today, he invents the following function:
Yousef will give you a list of queries, and you need to find the answers for them. For each query, you are given an integer n, and your task is to count the number of integers m in which (0 ≤ m ≤ n) and calc(n, m) is an even number. Can you?
Input
The first line contains an integer T (1 ≤ T ≤ 105) specifying the number of test cases.
Each test case consists of a single line containing an integer n (0 ≤ n ≤ 1018), as described in the problem statement above.
Output
For each test case, print a single line containing the number of integers m in which (0 ≤ m ≤ n) and calc(n, m) is an even number.
Example
Input
2 1 2
Output
0 1
题意:C[i][j] j<=i 的偶数有多少个
题解:打表偶数的找不到规律,那就看看奇数的啊,发现和2进制为1的个数有关
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=1e5+100;
typedef long long ll;
const ll mod=1e9+7;
ll C[1100][1100];
int main()
{
/*
C[0][0]=C[1][0]=C[1][1]=1;
for(int i=2;i<=1000;i++)
{
C[i][0]=1;
int cnt=1;
for(int j=1;j<=i;j++)
{
C[i][j]=(C[i-1][j]+C[i-1][j-1])%1000000;
if(C[i][j]&1) cnt++;
}
printf("%d: %d\n",i,cnt);
}
*/
int T;
scanf("%d",&T);
while(T--)
{
ll n,m,cnt=0;
scanf("%lld",&n);
m=n;
while(m)
{
cnt++;
m-=m&(-m);
}
printf("%lld\n",n+1-(1ll<<cnt));
}
return 0;
}