BZOJ3329: Xorequ(二进制数位dp 矩阵快速幂)
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2022-07-02 13:39:21
题意 "题目链接" Sol 挺套路的一道题 首先把式子移一下项 $x \oplus 2x = 3x$ 有一件显然的事情:$a \oplus b \leqslant c$ 又因为$a \oplus b + 2(a \& b) = c$ 那么$x \& 2x = 0$ 也就是说,$x$的二进制表示下不能 ......
题意
sol
挺套路的一道题
首先把式子移一下项
\(x \oplus 2x = 3x\)
有一件显然的事情:\(a \oplus b \leqslant c\)
又因为\(a \oplus b + 2(a \& b) = c\)
那么\(x \& 2x = 0\)
也就是说,\(x\)的二进制表示下不能有相邻位
第一问直接数位dp即可
第二问比较interesting,设\(f[i]\)表示二进制为\(i\)的方案数,转移时考虑上一位选不选
如果能选,方案数为\(f[i - 2]\)
不选的方案数为\(f[i - 1]\)
#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\n", 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\n", (a.m[1][1] + a.m[1][2]) % mod); } return 0; } /* 1 5 */