BZOJ3329 Xorequ（数位dp+矩阵快速幂）

　　显然当x中没有相邻的1时该式成立，看起来这也是必要的。

　　于是对于第一问，数位dp即可。第二问写出dp式子后发现就是斐波拉契数列，矩阵快速幂即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define P 1000000007
#define ll long long
int T,num[];
ll n,dp[][][];
struct matrix
{
    int n,a[][];
    matrix operator *(const matrix&b) const
    {
        matrix c;c.n=n;memset(c.a,,sizeof(c.a));
        for (register int i=;i<n;i++)
            for (register int j=;j<;j++)
                for (register int k=;k<;k++)
                c.a[i][j]=(c.a[i][j]+1ll*a[i][k]*b.a[k][j]%P)%P;
        return c;
    }
}f,a;
ll solve1(ll n)
{
    memset(dp,,sizeof(dp));
    int m=-;
    while (n) num[++m]=n&,n>>=;
    dp[m+][][]=;
    for (int i=m;~i;i--)
    if (num[i])
    {
        dp[i][][]=dp[i+][][]+dp[i+][][]+dp[i+][][]+dp[i+][][];
        dp[i][][]=dp[i+][][];
        dp[i][][]=dp[i+][][];
    }
    else
    {
        dp[i][][]=dp[i+][][]+dp[i+][][];
        dp[i][][]=dp[i+][][]+dp[i+][][];
        dp[i][][]=dp[i+][][];
    }
    return dp[][][]+dp[][][]+dp[][][]+dp[][][];
}
int solve2(ll n)
{
    a.n=;a.a[][]=;a.a[][]=a.a[][]=a.a[][]=;
    f.n=;f.a[][]=,f.a[][]=;
    for (;n;n>>=,a=a*a) if (n&) f=f*a;
    return f.a[][];
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj3329.in","r",stdin);
    freopen("bzoj3329.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    T=read();
    while (T--)
    {
        scanf(LL,&n);
        printf(LL,solve1(n)-);
        printf("%d\n",solve2(n+));
    }
    return ;
}