Maximum Flow 规律题 2017 ACM-ICPC 亚洲区（西安赛区）网络赛

Given a directed graph with nn nodes,
labeled 0,1,
\cdots, n-10,1,⋯,n−1.
For each <i,
j><i,j> satisfies 0
\le i < j < n0≤i<j<n,
there exists an edge from the i-th node to the j-th node, the capacity of which is ii xor jj.

Find the maximum flow network from the 0-th node to the (n-1)-th node, modulo 10000000071000000007.

Input Format

Multiple test cases (no more than 1000010000).

In each test case, one integer in a line denotes n(2
\le n \le 10^{18})n(2≤n≤10​18​​).

Output Format

Output the maximum flow modulo 10000000071000000007for
each test case.

样例输入

2

样例输出

1

完完全全找的规律,累死

先打了个表,发现其相邻数之差是有规律的

然后就各种调....

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MOD = 1000000007;

ll n;
ll er[1000];
ll er1[66];
ll ant[1000];

int main()
{
er[0] = 1;
er1[0] =1 ;
for(int i=1;i<=129;i++)
{
er[i] = 2*er[i-1]%MOD;
}

for(int i=1;i<=64;i++)
{
er1[i] = er1[i-1]*2;
}

ant[0] = 2;
for(int i=1;i<=64;i++)
{
ant[i] = er[i*2]+1;
ant[i]%=MOD;
}

ll tt = ant[0];
ll pr;
for(int i=1;i<=64;i++)
{
pr= ant[i];
ant[i] = (ant[i]+MOD-tt)%MOD;
tt = pr;
// cout <<ant[i]<<endl;
}
// freopen("data.txt","r",stdin);
while(cin >> n)
{
if(n==2)
{
cout<<1<<endl;
continue;
}
ll ans = 0;
ll t;
for(int i=0;i<=64;i++)
{
if(n<er1[i])
break;
if(n%er1[i]==0)
{
t= n/er1[i];
}
else
{
t= n/er1[i]+1;
}
t-=2;
if(t<=0)
{
break;
}
// cout <<t<<endl;
t%=MOD;
ans += t*ant[i];
ans%=MOD;
}
cout << ans+1<<endl;
}
return 0;
}