BZOJ3329: Xorequ(二進位制數位dp 矩陣快速冪)

自為風月馬前卒發表於2018-10-07
矩陣

題意

題目連結

Sol

挺套路的一道題

首先把式子移一下項

(x oplus 2x = 3x)

有一件顯然的事情：(a oplus b leqslant c)

又因為(a oplus b + 2(a & b) = c)

那麼(x & 2x = 0)

也就是說，(x)的二進位制表示下不能有相鄰位

第一問直接數位dp即可

第二問比較interesting，設(f[i])表示二進位制為(i)的方案數，轉移時考慮上一位選不選

如果能選，方案數為(f[i – 2])

不選的方案數為(f[i – 1])

#include<bits/stdc++.h>
#define LL long long 
//#define int long long 
#define file {freopen("a.in", "r", stdin); freopen("a.out", "w", stdout);}
using namespace std;
const int MAXN = 233, mod = 1e9 + 7;
inline LL read() {
    char c = getchar(); LL x = 0, f = 1;
    while(c < `0` || c > `9`) {if(c == `-`) f = -1; c = getchar();}
    while(c >= `0` && c <= `9`) x = x * 10 + c - `0`, c = getchar();
    return x * f;
}
LL N;
struct Matrix {
    int m[3][3];
    Matrix() {
        memset(m, 0, sizeof(m));
    }
    Matrix operator * (const Matrix &rhs) const {
        Matrix ans;
        for(int k = 1; k <= 2; k++)
            for(int i = 1; i <= 2; i++)
                for(int j = 1; j <= 2; j++)
                    (ans.m[i][j] += 1ll * m[i][k] * rhs.m[k][j] % mod) %= mod;
        return ans;
    }
};
Matrix MatrixPow(Matrix a, LL p) {
    Matrix base;
    for(int i = 1; i <= 2; i++) base.m[i][i] = 1;
    while(p) {
        if(p & 1) base = base * a;
        a = a * a; p >>= 1;
    }
    return base;
}
LL num[MAXN], tot; LL f[MAXN][2];
LL dfs(int x, bool lim, bool pre) {
    if(!lim && (~f[x][pre])) return f[x][pre];
    if(x == 0) return 1;
    LL ans = 0;
    if(!pre && (num[x] == 1 || (!lim))) ans += dfs(x - 1, lim, 1);
    ans += dfs(x - 1, lim && num[x] == 0, 0);

    if(!lim) f[x][pre] = ans;
    return ans;
}
LL dp(LL x) {
    tot = 0;
    while(x) num[++tot] = x & 1, x >>= 1;
    return dfs(tot, 1, 0);
}   
main() {
//  file;
    memset(f, -1, sizeof(f));
    int T = read();
    while(T--) {
        N = read();
        printf("%lld
", dp(N) - 1);
        Matrix a;
        a.m[1][1] = 1; a.m[1][2] = 1;
        a.m[2][1] = 1; a.m[2][2] = 0;
        a = MatrixPow(a, N);
        printf("%d
", (a.m[1][1] + a.m[1][2]) % mod);      
    }

    return 0;
}
/*
1
5
*/

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