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题意：两个人玩游戏，给定一个日期，他们轮流选择日期，可以选择当前日期的下一天，如果下一个月也有这一天的话则也可以选择下一个月的这一天。超过某一日期的人输。
思路：以天为状态，则一共有300多万个左右的状态，然后用dp预处理每个状态是必胜态还是必败态。一个状态是必胜态当且仅存在它的一个后继是必败态，一个状态是必败态当且仅当它的所有后继都是必胜态，写成逻辑式则是：state(now)= !state(next1) || !state(next2) || .. || !state(nextn)
由于空间不足，采用一个4位int存年月日，方法是用8个十进制位，高4位存年，中2位存月，低2位存日。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 | /* ******************************************************************************** */#include <iostream> //#include <cstdio> //#include <cmath> //#include <cstdlib> //#include <cstring> //#include <vector> //#include <ctime> //#include <deque> //#include <queue> //#include <algorithm> //#include <map> //#include <cmath> //using namespace std; // //#define pb push_back //#define mp make_pair //#define X first //#define Y second //#define all(a) (a).begin(), (a).end() //#define fillchar(a, x) memset(a, x, sizeof(a)) // //typedef pair<int, int> pii; //typedef long long ll; //typedef unsigned long long ull; // //#ifndef ONLINE_JUDGE //void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);} //void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> //void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1; //while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> //void print(const T t){cout<<t<<endl;}template<typename F,typename...R> //void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> //void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} //#endif // ONLINE_JUDGE //template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);} //template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);} //template<typename T> //void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];} //template<typename T> //void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];} // //const double PI = acos(-1.0); //const int INF = 1e9 + 7; // ///* -------------------------------------------------------------------------------- */const int month[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int maxn = 4e6;struct Node { int s; int gety() { return s / 10000; } int getm() { return s / 100 % 100; } int getd() { return s % 100; } Node(int y = 0, int m = 0, int d = 0) { s = y * 10000 + m * 100 + d; } Node nextm() { int y = gety(), m = getm() + 1, d = getd(); if (m == 12) { y ++; m = 0; } return Node(y, m, d); }};Node state[maxn];int cnt;int hsh[102][12][31];bool dp[maxn];bool leap(int y) { return y % 4 == 0 && y % 100 != 0 || y % 400 == 0;}void pre_init() { for (int i = 0; i < 102; i ++) { for (int j = 0; j < 12; j ++) { int M = month[j]; if (leap(i + 1900) && j == 1) M ++; for (int k = 0; k < M; k ++) { state[cnt] = Node(i, j, k); hsh[i][j][k] = cnt ++; } } } for (int i = hsh[101][10][3] + 1; i < cnt; i ++) dp[i] = true; for (int i = hsh[101][10][3] - 1; i >= 0; i --) { Node S = state[i].nextm(); int y = S.gety(), m = S.getm(), d = S.getd(); dp[i] = !dp[i + 1]; if (hsh[y][m][d]) dp[i] = dp[i] || !dp[hsh[y][m][d]]; }}int main() {#ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout);#endif // ONLINE_JUDGE pre_init(); int T; cin >> T; while (T --) { int y, m, d; scanf("%d%d%d", &y, &m, &d); int n = hsh[y - 1900][m - 1][d - 1]; puts(dp[n]? "YES" : "NO"); } return 0;}/* ******************************************************************************** */ |
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原文地址：http://www.cnblogs.com/jklongint/p/4700298.html
