bzoj3329: Xorequ 数位DP

题目链接

bzoj3329: Xorequ

题解

x^3x=2x
x^2x=3x
因为
a^b+((a&b)<<1)=a+b
x^2x=x+2x
那么x和2x的二进制表示中不存在相邻的1
对于第一问数位dp
第二问写出现递推式矩乘优化一下

代码

#include<cstdio> 
#include<cstring> 
#include<algorithm> 
#define rep(p,x,k) for(int p = x;p <= k;++ p) 
#define per(p,x,k) for(int p = x;p >= k;-- p) 
#define gc getchar()
#define pc putchar 
#define LL long long  
inline LL read() {
    LL x = 0,f = 1; 
    char c = gc; 
    while(c < '0' || c > '9') c = gc; 
    while(c <= '9' && c >= '0') x = x * 10 + c -'0',c = gc; 
    return x ; 
} 
void print(LL x) { 
    if(x < 0) { 
        pc('-'); 
        x = -x; 
    } 
    if(x >= 10) print(x / 10); 
    pc(x % 10 + '0'); 
} 
const int maxn = 107; 
const int  mod = 1e9 + 7;  
struct Matrix { 
    LL a[3][3]; 
    Matrix() {memset(a,0,sizeof a); }  
    Matrix operator * (const Matrix &t) const{ 
        Matrix ret; 
        rep(k,0,1) 
            rep(i,0,1) 
                rep(j,0,1) { 
                    (ret.a[i][j] += a[i][k] * t.a[k][j] % mod) %= mod; 
                } 
        return ret; 
    } 
} ; 
int n,m; 
int bit[100]; 
LL f[107][3]; 
LL dfs(int x,int lim,int las) { 
    if(!x) return 1; 
    if(!lim && f[x][las]) return f[x][las];  
    LL ret = 0; 
    int ulim = lim ? bit[x] : 1; 
    ret += dfs(x - 1,lim && !ulim,0); 
    if(ulim && !las) ret += dfs(x - 1,lim,1); 
    ret; 
    if(!lim) f[x][las] = ret;  
    return ret; 
} 
LL solve(LL n) { 
    int cnt = 0; 
    while(n) { 
        if(n & 1) bit[++ cnt] = 1; 
        else bit[++ cnt] = 0; 
        n >>= 1; 
    } 
    return dfs(cnt,1,0) - 1; 
} 
LL fstpow(Matrix a,LL k) { 
    Matrix ret; 
    ret.a[0][0] = ret.a[0][1] = 1; 
    for(;k;k >>= 1,a = a * a)  
        if(k & 1) ret = ret * a;  
    return ret.a[0][0] ;  
}  
int main() {  
    //freopen("1.in","r",stdin); freopen("A.out","w",stdout); 
    Matrix a;  
    a.a[0][0] = a.a[0][1] = a.a[1][0] = 1;  
    int T = read();  
    rep(i,1,T) { 
        LL n; 
        n = read(); 
        print(solve(n)); 
        pc('\n'); 
        print(fstpow(a,n)); 
        pc('\n'); 
    }   
    return 0; 
}