Brief description:
给定一个 n 个元素的数组 {ai}, 问存在多少组 {xi, xi <= ai && 所有 xi 的 xor 和 等于 0}。。。 ( n <= 50 .. )
亦或和等于 0 和等于 k 的版本几乎没有区别,讨论前者。
这类题目的解法是从高到低按位分阶段组合 DP .. .
我们的目的是,在每轮的 DP 过后,这一位上所有至少 Skip 一位的组合,全部被计数,这样做之后,剩下的所有还没有被计数的情况(没有发生 Skip)就会形成一个结构相同的子问题 …
额,还没有定义 Skip …
我们讨论当前扫描的是最高位 … 对于那些在这一位上是 1 的元素,这些元素有两种决策:
分别 是取 1 和取 0,我们称那些含有 1 的元素,取 0 的决策为 折越 (Skip) …
那么对于当前第 i 位,扫描了 j 个数字,我们设计状态 f[0], f[1], f[2] 分别为
从未发生过 折越 、发生过奇数次 折越、和发生过偶数次 折越 且不包括 f[0]。
分类讨论 dp 。。。
#define c ((a[j] & _U(i)) + 1) #define g (1 << i) … REP(j, n){ if (_1(a[j], i)) { int w1 = sum(f[0], pdt(f[2], g)), w2 = pdt(f[1], g); t ^= 1; MUL(f[0], c), MUL(f[1], c), MUL(f[2], c); INC(f[1], w1), INC(f[2], w2); } else { MUL(f[0], c), MUL(f[1], c), MUL(f[2], c); } } …
之后,我们考虑含有 1 的个数的奇偶性,
如果含有奇数个1,那么我们前面所说没有发生过一次 折越 (Skip) 的情况就不会出现在计数方案中,因此直接跳出。
每一位的复杂度为 O(n) ,总的复杂度为 O(nb),b 表元素的二进制位数 … (最后边界处别忘了所有元素都为 0 的话也是一组解 .. .)
…
.
.
(另,推公式的过程中还有一个难点,具体的说来是这样,对于当前扫描到第 i 位,一个发生了 k 次 折越 的状态,是需要保证 (i, 0] 位上的亦或和为 0 的。
因为发生 折越 的位置之后的二进制位可以任意组合,所以如果固定了其中的任意 k-1 个元素的状态,总可以设置一种第 k 个元素的状态,使得 (i, 0] 位上的亦或和为 0,这也是组合 dp 可以奏效的依据。)
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/* …………………………………………………………………………………………………………………. */ | |
/** Header **/ | |
#include <iostream> | |
#include <iomanip> | |
#include <sstream> | |
#include <string> | |
#include <vector> | |
#include <cstdio> | |
#include <cstring> | |
#include <cmath> | |
#include <algorithm> | |
#include <bitset> | |
#include <list> | |
#include <set> | |
#include <map> | |
using namespace std; | |
/** Repeat **/ | |
#define REP(I, N) for (int I=0;I<int(N);++I) | |
#define FOR(I, A, B) for (int I=int(A);I<int(B);++I) | |
#define DWN(I, B, A) for (int I=int(B-1);I>=int(A);–I) | |
#define REP_1(I, N) for (int I=1;I<=int(N);++I) | |
#define FOR_1(I, A, B) for (int I=int(A);I<=int(B);++I) | |
#define DWN_1(I, B, A) for (int I=int(B);I>=int(A);–I) | |
#define REP_C(I, N) for (int N____=int(N),I=0;I<N____;++I) | |
#define FOR_C(I, A, B) for (int B____=int(B),I=A;I<B____;++I) | |
#define DWN_C(I, B, A) for (int A____=int(A),I=B-1;I>=A____;–I) | |
#define REP_1_C(I, N) for (int N____=int(N),I=1;I<=N____;++I) | |
#define FOR_1_C(I, A, B) for (int B____=int(B),I=A;I<=B____;++I) | |
#define DWN_1_C(I, B, A) for (int A____=int(A),I=B;I>=A____;–I) | |
#define DO(N) while(N–) | |
#define DO_C(N) int N____ = N; while(N____–) | |
#define TO(i, a, b) int s_=a<b?1:-1,b_=b+s_;for(int i=a;i!=b_;i+=s_) | |
#define TO_1(i, a, b) int s_=a<b?1:-1,b_=b;for(int i=a;i!=b_;i+=s_) | |
#define SQZ(I, J, A, B) for (int I=int(A),J=int(B)-1;I<J;++I,–J) | |
#define SQZ_1(I, J, A, B) for (int I=int(A),J=int(B);I<=J;++I,–J) | |
/** Micro Mezzo Macro Flation — Overheated Economy **/ | |
#define ALL(A) A.begin(), A.end() | |
#define CPY(A, B) memcpy(A, B, sizeof(A)) | |
#define INS(A, P, B) A.insert(A.begin() + P, B) | |
#define ERS(A, P) A.erase(A.begin() + P) | |
#define SRT(A) sort(ALL(A)) | |
#define BSC(A, X) find(ALL(A), X) // != A.end() | |
#define SZ(A) int(A.size()) | |
#define PB push_back | |
#define MP(A, B) make_pair(A, B) | |
#define Rush int T____; RD(T____); DO(T____) | |
#pragma comment(linker, "/STACK:36777216") | |
#define Ruby system("ruby main.rb") | |
#define Haskell system("runghc main.hs") | |
#define Pascal system("fpc main.pas") | |
/** Typedef **/ | |
typedef long long LL; | |
typedef double DB; | |
typedef unsigned UINT; | |
typedef unsigned long long ULL; | |
typedef vector<int> VI; | |
typedef vector<string> VS; | |
typedef vector<LL> VL; | |
typedef vector<DB> VD; | |
typedef set<int> SI; | |
typedef set<string> SS; | |
typedef set<LL> SL; | |
typedef set<DB> SD; | |
typedef map<int, int> MI; | |
typedef map<string, int> MS; | |
typedef map<LL, int> ML; | |
typedef map<DB, int> MD; | |
typedef pair<int, int> PII; | |
typedef pair<int, bool> PIB; | |
typedef vector<PII> VII; | |
typedef set<PII> SII; | |
typedef map<PII, int> MII; | |
typedef vector<VI> VVI; | |
typedef vector<VII> VVII; | |
/** I/O Accelerator **/ | |
/* … :" We are I/O Accelerator … Use us at your own risk 😉 … " .. */ | |
template<class T> inline void RD(T &x){ | |
cin >> x; | |
//scanf("%d", &x); | |
//char c; for (c = getchar(); c < '0'; c = getchar()); x = c – '0'; for (c = getchar(); c >= '0'; c = getchar()) x = x * 10 + c – '0'; | |
//char c; c = getchar(); x = c – '0'; for (c = getchar(); c >= '0'; c = getchar()) x = x * 10 + c – '0'; | |
} | |
template<class T> inline void OT(const T &x){ | |
cout << x << endl; | |
//printf("%d\n", x); | |
//printf("%.2lf", x); | |
//printf("Case %d: %d", ____T, ans) | |
} | |
template<class T> inline T& _RD(T &x){ RD(x); return x;} | |
inline int RD(){ int x; RD(x); return x;} | |
inline LL RD_LL(){ int x; RD(x); return x;} | |
inline DB RD_DB(){ DB x; RD(x); return x;} | |
template<class T0, class T1> inline void RD(T0 &x0, T1 &x1){RD(x0), RD(x1);} | |
template<class T0, class T1, class T2> inline void RD(T0 &x0, T1 &x1, T2 &x2){RD(x0), RD(x1), RD(x2);} | |
template<class T0, class T1, class T2, class T3> inline void RD(T0 &x0, T1 &x1, T2 &x2, T3 &x3){RD(x0), RD(x1), RD(x2), RD(x3);} | |
template<class T0, class T1, class T2, class T3, class T4> inline void RD(T0 &x0, T1 &x1, T2 &x2, T3 &x3, T4 &x4){RD(x0), RD(x1), RD(x2), RD(x3), RD(x4);} | |
template<class T0, class T1, class T2, class T3, class T4, class T5> inline void RD(T0 &x0, T1 &x1, T2 &x2, T3 &x3, T4 &x4, T5 &x5){RD(x0), RD(x1), RD(x2), RD(x3), RD(x4), RD(x5);} | |
template<class T0, class T1, class T2, class T3, class T4, class T5, class T6> inline void RD(T0 &x0, T1 &x1, T2 &x2, T3 &x3, T4 &x4, T5 &x5, T6 &x6){RD(x0), RD(x1), RD(x2), RD(x3), RD(x4), RD(x5), RD(x6);} | |
template<class T0, class T1> inline void OT(T0 &x0, T1 &x1){OT(x0), OT(x1);} | |
template<class T0, class T1, class T2> inline void OT(T0 &x0, T1 &x1, T2 &x2){OT(x0), OT(x1), OT(x2);} | |
template<class T0, class T1, class T2, class T3> inline void OT(T0 &x0, T1 &x1, T2 &x2, T3 &x3){OT(x0), OT(x1), OT(x2), OT(x3);} | |
template<class T0, class T1, class T2, class T3, class T4> inline void OT(T0 &x0, T1 &x1, T2 &x2, T3 &x3, T4 &x4){OT(x0), OT(x1), OT(x2), OT(x3), OT(x4);} | |
template<class T0, class T1, class T2, class T3, class T4, class T5> inline void OT(T0 &x0, T1 &x1, T2 &x2, T3 &x3, T4 &x4, T5 &x5){OT(x0), OT(x1), OT(x2), OT(x3), OT(x4), OT(x5);} | |
template<class T0, class T1, class T2, class T3, class T4, class T5, class T6> inline void OT(T0 &x0, T1 &x1, T2 &x2, T3 &x3, T4 &x4, T5 &x5, T6 &x6){OT(x0), OT(x1), OT(x2), OT(x3), OT(x4), OT(x5), OT(x6);} | |
template<class T> inline void RST(T &A){memset(A, 0, sizeof(A));} | |
template<class T0, class T1> inline void RST(T0 &A0, T1 &A1){RST(A0), RST(A1);} | |
template<class T0, class T1, class T2> inline void RST(T0 &A0, T1 &A1, T2 &A2){RST(A0), RST(A1), RST(A2);} | |
template<class T0, class T1, class T2, class T3> inline void RST(T0 &A0, T1 &A1, T2 &A2, T3 &A3){RST(A0), RST(A1), RST(A2), RST(A3);} | |
template<class T0, class T1, class T2, class T3, class T4> inline void RST(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4){RST(A0), RST(A1), RST(A2), RST(A3), RST(A4);} | |
template<class T0, class T1, class T2, class T3, class T4, class T5> inline void RST(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, T5 &A5){RST(A0), RST(A1), RST(A2), RST(A3), RST(A4), RST(A5);} | |
template<class T0, class T1, class T2, class T3, class T4, class T5, class T6> inline void RST(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, T5 &A5, T6 &A6){RST(A0), RST(A1), RST(A2), RST(A3), RST(A4), RST(A5), RST(A6);} | |
template<class T> inline void CLR(T &A){A.clear();} | |
template<class T0, class T1> inline void CLR(T0 &A0, T1 &A1){CLR(A0), CLR(A1);} | |
template<class T0, class T1, class T2> inline void CLR(T0 &A0, T1 &A1, T2 &A2){CLR(A0), CLR(A1), CLR(A2);} | |
template<class T0, class T1, class T2, class T3> inline void CLR(T0 &A0, T1 &A1, T2 &A2, T3 &A3){CLR(A0), CLR(A1), CLR(A2), CLR(A3);} | |
template<class T0, class T1, class T2, class T3, class T4> inline void CLR(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4){CLR(A0), CLR(A1), CLR(A2), CLR(A3), CLR(A4);} | |
template<class T0, class T1, class T2, class T3, class T4, class T5> inline void CLR(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, T5 &A5){CLR(A0), CLR(A1), CLR(A2), CLR(A3), CLR(A4), CLR(A5);} | |
template<class T0, class T1, class T2, class T3, class T4, class T5, class T6> inline void CLR(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, T5 &A5, T6 &A6){CLR(A0), CLR(A1), CLR(A2), CLR(A3), CLR(A4), CLR(A5), CLR(A6);} | |
template<class T> inline void CLR(T &A, int n){REP(i, n) CLR(A[i]);} | |
template<class T> inline void FLC(T &A, int x){memset(A, x, sizeof(A));} | |
template<class T0, class T1> inline void FLC(T0 &A0, T1 &A1, int x){FLC(A0, x), FLC(A1, x);} | |
template<class T0, class T1, class T2> inline void FLC(T0 &A0, T1 &A1, T2 &A2){FLC(A0), FLC(A1), FLC(A2);} | |
template<class T0, class T1, class T2, class T3> inline void FLC(T0 &A0, T1 &A1, T2 &A2, T3 &A3){FLC(A0), FLC(A1), FLC(A2), FLC(A3);} | |
template<class T0, class T1, class T2, class T3, class T4> inline void FLC(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4){FLC(A0), FLC(A1), FLC(A2), FLC(A3), FLC(A4);} | |
template<class T0, class T1, class T2, class T3, class T4, class T5> inline void FLC(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, T5 &A5){FLC(A0), FLC(A1), FLC(A2), FLC(A3), FLC(A4), FLC(A5);} | |
template<class T0, class T1, class T2, class T3, class T4, class T5, class T6> inline void FLC(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, T5 &A5, T6 &A6){FLC(A0), FLC(A1), FLC(A2), FLC(A3), FLC(A4), FLC(A5), FLC(A6);} | |
/** Add – On **/ | |
const int MOD = 1000000003; | |
const int INF = 0x7fffffff; | |
template<class T> inline void checkMin(T &a, T b){if (b<a) a=b;} | |
template<class T> inline void checkMax(T &a, T b){if (b>a) a=b;} | |
template<class T> inline T sqr(T x){return x * x;} | |
inline bool _1(int x, int i){return x & 1<<i;} | |
inline int _1(int i){return 1<<i;} | |
inline int _U(int i){return _1(i) – 1;}; | |
/* …………………………………………………………………………………………………………………. */ | |
const int N = 50; | |
UINT a[N]; ULL ans; | |
int n, k; | |
#define c ((a[j] & _U(i)) + 1) | |
#define g (1 << i) | |
void solve() { | |
ULL f[3]; int t; DWN(i, 32 ,0) { | |
t = 0, f[0] = 1, f[1] = 0, f[2] = 0; //# | |
REP(j, n){ | |
if (_1(a[j], i)) { | |
ULL w1 = f[0] + f[2] * g, w2 = f[1] * g; t ^= 1; | |
f[0] *= c, f[1] *= c, f[2] *= c; | |
f[1] += w1, f[2] += w2; | |
} | |
else { | |
f[0] *= c, f[1] *= c, f[2] *= c; | |
} | |
} | |
if (t) { | |
ans += f[1] – 1; | |
return; | |
} | |
ans += f[2]; | |
} | |
} | |
int main() { | |
//freopen("D.in", "r", stdin); | |
//freopen("D.out", "w", stdout); | |
//ios::sync_with_stdio(false); | |
RD(n); REP(i, n) RD(a[i]); | |
solve(), OT(ans); | |
} |
http://www.lydsy.com/JudgeOnline/problem.php?id=1442
http://livearchive.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=384&problem=3102&mosmsg=Submission+received+with+ID+856655
http://main.edu.pl/en/archive/oi/13/kry
SRM 508 DIV 1 Level 2. YetAnotherORProblem
http://hi.baidu.com/%D2%BB%CE%BB%C1%E3/blog/item/603ee5b68bf283e630add144.html
http://blog.163.com/fjxmlhx@126/blog/static/144072841201022492458853/