bzoj3329: Xorequ 數位DP
阿新 • • 發佈：2018-10-17
class bit a.out include turn a+b -- cst etc
題目鏈接
bzoj3329: Xorequ
題解
x^3x=2x
x^2x=3x
因為
a^b+((a&b)<<1)=a+b
x^2x=x+2x
那麽x和2x的二進制表示中不存在相鄰的1
對於第一問數位dp
第二問寫出現遞推式矩乘優化一下
代碼
#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; }
bzoj3329: Xorequ 數位DP