2015百度之星资格赛题解

题目链接： http://bestcoder.hdu.edu.cn/contests/contest_show.php?cid=584

A.

思路：题目可以看成这样一个模型，把n个物品分成若干组，使得每组只有1个或两个物品的方法总数。令dp[n]表示n个物品的答案，则有:

dp[n] = dp[n-1] + (n - 1) * dp[n - 2]

代码如下： 

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#pragma comment(linker, "/STACK:10240000,10240000") #include <iostream> #include <cstdio> #include <algorithm> #include <cstdlib> #include <cstring> #include <map> #include <queue> #include <deque> #include <cmath> #include <vector> #include <ctime> #include <cctype> #include <set> #include <bitset> #include <functional> #include <numeric> #include <stdexcept> #include <utility> using namespace std; #define mem0(a) memset(a, 0, sizeof(a)) #define mem_1(a) memset(a, -1, sizeof(a)) #define lson l, m, rt << 1 #define rson m + 1, r, rt << 1 | 1 #define rep_up0(a, b) for (int a = 0; a < (b); a++) #define rep_up1(a, b) for (int a = 1; a <= (b); a++) #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--) #define rep_down1(a, b) for (int a = b; a > 0; a--) #define all(a) (a).begin(), (a).end() #define lowbit(x) ((x) & (-(x))) #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {} #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} #define pchr(a) putchar(a) #define pstr(a) printf("%s", a) #define sstr(a) scanf("%s", a) #define sint(a) scanf("%d", &a) #define sint2(a, b) scanf("%d%d", &a, &b) #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c) #define pint(a) printf("%d\n", a) #define test_print1(a) cout << "var1 = " << a << endl #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl #define mp(a, b) make_pair(a, b) #define pb(a) push_back(a) typedef unsigned int uint; typedef long long LL; typedef pair< int , int > pii; typedef vector< int > vi; const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1}; const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 }; const int maxn = 1e6 + 7; const int md = 1000000007; const int inf = 1e9 + 7; const LL inf_L = 1e18 + 7; const double pi = acos (-1.0); const double eps = 1e-6; template < class T>T gcd(T a, T b){ return b==0?a:gcd(b,a%b);} template < class T> bool max_update(T &a, const T &b){ if (b>a){a = b; return true ;} return false ;} template < class T> bool min_update(T &a, const T &b){ if (b<a){a = b; return true ;} return false ;} template < class T>T condition( bool f, T a, T b){ return f?a:b;} template < class T> void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];} int make_id( int x, int y, int n) { return x * n + y; } int ans[maxn]; void init() { ans[1] = 1; ans[2] = 2; for ( int i = 3; i <= 1000000; i ++) { ans[i] = (ans[i - 1] + (LL)(i - 1) * ans[i - 2]) % md; } } int main() { //freopen("in.txt", "r", stdin); init(); int T, n; cin >> T; rep_up1(cas, T) { cin >> n; printf ( "Case #%d:\n%d\n" , cas, ans[n]); } return 0; }

 

B.

思路：直接还原原矩阵或者找输入输出的对应关系即可 。

代码如下：

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#pragma comment(linker, "/STACK:10240000,10240000") #include <iostream> #include <cstdio> #include <algorithm> #include <cstdlib> #include <cstring> #include <map> #include <queue> #include <deque> #include <cmath> #include <vector> #include <ctime> #include <cctype> #include <set> #include <bitset> #include <functional> #include <numeric> #include <stdexcept> #include <utility> using namespace std; #define mem0(a) memset(a, 0, sizeof(a)) #define mem_1(a) memset(a, -1, sizeof(a)) #define lson l, m, rt << 1 #define rson m + 1, r, rt << 1 | 1 #define rep_up0(a, b) for (int a = 0; a < (b); a++) #define rep_up1(a, b) for (int a = 1; a <= (b); a++) #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--) #define rep_down1(a, b) for (int a = b; a > 0; a--) #define all(a) (a).begin(), (a).end() #define lowbit(x) ((x) & (-(x))) #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {} #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} #define pchr(a) putchar(a) #define pstr(a) printf("%s", a) #define sstr(a) scanf("%s", a) #define sint(a) scanf("%d", &a) #define sint2(a, b) scanf("%d%d", &a, &b) #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c) #define pint(a) printf("%d\n", a) #define test_print1(a) cout << "var1 = " << a << endl #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl #define mp(a, b) make_pair(a, b) #define pb(a) push_back(a) typedef unsigned int uint; typedef long long LL; typedef pair< int , int > pii; typedef vector< int > vi; const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1}; const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 }; const int maxn = 120000; const int md = 1000000007; const int inf = 1e9 + 7; const LL inf_L = 1e18 + 7; const double pi = acos (-1.0); const double eps = 1e-6; template < class T>T gcd(T a, T b){ return b==0?a:gcd(b,a%b);} template < class T> bool max_update(T &a, const T &b){ if (b>a){a = b; return true ;} return false ;} template < class T> bool min_update(T &a, const T &b){ if (b<a){a = b; return true ;} return false ;} template < class T>T condition( bool f, T a, T b){ return f?a:b;} template < class T> void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];} int make_id( int x, int y, int n) { return x * n + y; } vector<vector< char > > ch; char s[maxn]; int main() { //freopen("in.txt", "r", stdin); int T, k; cin >> T; scanf ( "%*c" ); rep_up1(cas, T) { gets (s); scanf ( "%d%*c" , &k); printf ( "Case #%d:\n" , cas); int n = strlen (s); ch.resize(n / k + 2); rep_up0(i, n / k + 2) ch[i].resize(k); int cur = 0; rep_up0(i, n % k) { rep_up0(j, n / k + 1) { ch[j][i] = s[cur ++]; } } for ( int i = n % k; i < k; i ++) { rep_up0(j, n / k) { ch[j][i] = s[cur ++]; } } rep_up0(i, n / k) { rep_up0(j, k) { pchr(ch[i][j]); } } rep_up0(i, n % k) { pchr(ch[n / k][i]); } cout << endl; } return 0; }

 

C.

思路：二进制运算和map判重。

代码如下： 

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#pragma comment(linker, "/STACK:10240000,10240000") #include <iostream> #include <cstdio> #include <algorithm> #include <cstdlib> #include <cstring> #include <map> #include <queue> #include <deque> #include <cmath> #include <vector> #include <ctime> #include <cctype> #include <set> #include <bitset> #include <functional> #include <numeric> #include <stdexcept> #include <utility> using namespace std; #define mem0(a) memset(a, 0, sizeof(a)) #define mem_1(a) memset(a, -1, sizeof(a)) #define lson l, m, rt << 1 #define rson m + 1, r, rt << 1 | 1 #define rep_up0(a, b) for (int a = 0; a < (b); a++) #define rep_up1(a, b) for (int a = 1; a <= (b); a++) #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--) #define rep_down1(a, b) for (int a = b; a > 0; a--) #define all(a) (a).begin(), (a).end() #define lowbit(x) ((x) & (-(x))) #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {} #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} #define pchr(a) putchar(a) #define pstr(a) printf("%s", a) #define sstr(a) scanf("%s", a) #define sint(a) scanf("%d", &a) #define sint2(a, b) scanf("%d%d", &a, &b) #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c) #define pint(a) printf("%d\n", a) #define test_print1(a) cout << "var1 = " << a << endl #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl #define mp(a, b) make_pair(a, b) #define pb(a) push_back(a) typedef unsigned int uint; typedef long long LL; typedef pair< int , int > pii; typedef vector< int > vi; const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1}; const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 }; const int maxn = 120000; const int md = 1000000007; const int inf = 1e9 + 7; const LL inf_L = 1e18 + 7; const double pi = acos (-1.0); const double eps = 1e-6; template < class T>T gcd(T a, T b){ return b==0?a:gcd(b,a%b);} template < class T> bool max_update(T &a, const T &b){ if (b>a){a = b; return true ;} return false ;} template < class T> bool min_update(T &a, const T &b){ if (b<a){a = b; return true ;} return false ;} template < class T>T condition( bool f, T a, T b){ return f?a:b;} template < class T> void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];} int make_id( int x, int y, int n) { return x * n + y; } map<uint, bool > mp; uint a[1010]; int main() { //freopen("in.txt", "r", stdin); int T, w, x, y, z, n, m; cin >> T; rep_up1(cas, T) { printf ( "Case #%d:\n" , cas); cin >> n >> m; rep_up0(i, n) { scanf ( "%d.%d.%d.%d" , &w, &x, &y, &z); a[i] = (uint)w << 24 | x << 16 | y << 8 | z; } rep_up0(i, m) { scanf ( "%d.%d.%d.%d" , &w, &x, &y, &z); uint p = (uint)w << 24 | x << 16 | y << 8 | z; mp.clear(); int ans = 0; rep_up0(i, n) { uint res = p & a[i]; if (!mp[res]) { ans ++; mp[res] = 1; } } printf ( "%d\n" , ans); } } return 0; }

 

D.

思路：只要先手可以走第一步，那么它一定会走完最后一步，因为之后只需走后手的对称位置。

代码如下：