BZOJ3329 Xorequ 数位dp

 noi第一ysy大佬题解。

#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdio>
#include<iostream>

#define mod 1000000007
#define ll long long
using namespace std;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
    return x*f;
}

ll n;
ll f[64][2][2],bin[67];
ll a[3][3],b[3][3],ans[3][3],t[3][3],g[3];

void mul(ll b)
{
    ans[1][1]=1,ans[2][2]=1;ans[1][2]=ans[2][1]=0;
    while(b)
    {
        if(b&1)
        {
            memset(t,0,sizeof(t));
            for (int i=1;i<=2;i++)
                for (int j=1;j<=2;j++)
                    for (int k=1;k<=2;k++)
                        (t[i][j]+=ans[i][k]*a[k][j])%=mod;
            for (int i=1;i<=2;i++)
                for (int j=1;j<=2;j++)
                    ans[i][j]=t[i][j];
        }
        memset(t,0,sizeof(t));
        for (int i=1;i<=2;i++)
            for (int j=1;j<=2;j++)
                for (int k=1;k<=2;k++)
                    (t[i][j]+=a[i][k]*a[k][j])%=mod;
        for (int i=1;i<=2;i++)
            for (int j=1;j<=2;j++)
                a[i][j]=t[i][j];
        b>>=1;        
    }
}
int main()
{
    int T=read();
    bin[0]=1;
    for (ll i=1;i<=63;i++)
        bin[i]=bin[i-1]<<1;
    while(T--)
    {
        memset(f,0,sizeof(f));
        n=read();
        ll x=0;while(bin[x]<=n)x++;x--;
        f[x][1][1]=1,f[x][0][0]=1;
        for (ll i=x-1;i>=0;i--)
        {
            if(n&bin[i])
            {
                f[i][1][1]=f[i+1][0][1];
                f[i][1][0]=f[i+1][0][0];
                f[i][0][0]=f[i+1][1][1]+f[i+1][1][0]+f[i+1][0][1]+f[i+1][0][0];
            }
            else
            {
                f[i][1][0]=f[i+1][0][0];
                f[i][0][0]=f[i+1][1][0]+f[i+1][0][0];
                f[i][0][1]=f[i+1][1][1]+f[i+1][0][1];
            }
        }
        printf("%lld
",f[0][0][0]+f[0][0][1]+f[0][1][0]+f[0][1][1]-1);
        g[1]=2,g[2]=3;
        if(n<=2)
        {
            printf("%lld
",g[n]);
            continue;
        }
        a[1][1]=0,a[1][2]=1,a[2][1]=1,a[2][2]=1;
        mul(n-2);
        printf("%lld
",(ans[1][2]*2+ans[2][2]*3)%mod);
    }
}