SRM 358(1-250,500pt)

DIV1 250pt

题意：电视目前停留在第100台，有一个遥控器，可以向上或向下换台(需要按键一次)，也可以按一些数字，然后直接跳到该台(需要按键次数等于数字数，不需要按确定键)。但是，这个遥控一些数字键是坏的不能按。问要换到x台最少需要按多少次。x <= 500000。

解法：直接搜索。可能用bfs会快点，但我更喜欢写dfs就用了。

tag:search

 1 // BEGIN CUT HERE
 2 /*
 3 
 4 */
 5 // END CUT HERE
 6 #line 7 "BrokenButtons.cpp"
 7 #include <cstdlib>
 8 #include <cctype>
 9 #include <cstring>
10 #include <cstdio>
11 #include <cmath>
12 #include <algorithm>
13 #include <vector>
14 #include <iostream>
15 #include <sstream>
16 #include <set>
17 #include <queue>
18 #include <fstream>
19 #include <numeric>
20 #include <iomanip>
21 #include <bitset>
22 #include <list>
23 #include <stdexcept>
24 #include <functional>
25 #include <string>
26 #include <utility>
27 #include <map>
28 #include <ctime>
29 #include <stack>
30 
31 using namespace std;
32 
33 #define clr0(x) memset(x, 0, sizeof(x))
34 #define clr1(x) memset(x, -1, sizeof(x))
35 #define pb push_back
36 #define mp make_pair
37 #define sz(v) ((int)(v).size())
38 #define out(x) cout<<#x<<":"<<(x)<<endl
39 #define tst(x) cout<<x<<":"<<" "
40 #define tst1(a) cout<<#a<<endl
41 #define CINBEQUICKER std::ios::sync_with_stdio(false)
42 
43 typedef vector<int> vi;
44 typedef vector<string> vs;
45 typedef vector<double> vd;
46 typedef long long int64;
47 
48 const double eps = 1e-8;
49 const double PI = atan(1.0)*4;
50 const int inf = 2139062143 / 2;
51 
52 inline int MyMod( int a , int b ) { return (a%b+b)%b;}
53 
54 class BrokenButtons
55 {
56     public:
57         int ans, len, p;
58         bool v[20];
59         void dfs(int x, int num)
60         {
61             if (num) ans = min(ans, abs(p-x) + num);
62             if (num <= len)
63                 for (int i = 0; i < 10; ++ i) if (!v[i]) dfs (x*10+i, num+1);
64         }
65         int minPresses(int pag, vector <int> bro){
66             clr0 (v);
67             for (int i = 0; i < sz(bro); ++ i) v[bro[i]] = 1;
68             ans = abs(pag-100); p = pag; len = 0;
69             while (pag) pag /= 10, len ++;
70             dfs (0, 0);
71             return ans;
72         }
73         
74 // BEGIN CUT HERE
75     public:
76     void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); if ((Case == -1) || (Case == 4)) test_case_4(); }
77     private:
78     template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
79     void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
80     void test_case_0() { int Arg0 = 5457; int Arr1[] = { 6, 7, 8 }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 6; verify_case(0, Arg2, minPresses(Arg0, Arg1)); }
81     void test_case_1() { int Arg0 = 100; int Arr1[] = { 1, 0, 5 }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 0; verify_case(1, Arg2, minPresses(Arg0, Arg1)); }
82     void test_case_2() { int Arg0 = 14124; int Arr1[] = { }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 5; verify_case(2, Arg2, minPresses(Arg0, Arg1)); }
83     void test_case_3() { int Arg0 = 1; int Arr1[] = {1, 2, 3, 4, 5, 6, 7, 8, 9}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 2; verify_case(3, Arg2, minPresses(Arg0, Arg1)); }
84     void test_case_4() { int Arg0 = 80000; int Arr1[] = { 8, 9 }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 2228; verify_case(4, Arg2, minPresses(Arg0, Arg1)); }
85 
86 // END CUT HERE
87 
88 };
89 //by plum rain
90 // BEGIN CUT HERE
91 int main()
92 {
93     //freopen( "a.out" , "w" , stdout );    
94     BrokenButtons ___test;
95     ___test.run_test(-1);
96        return 0;
97 }
98 // END CUT HERE
View Code

 

DIV1 500pt

题意：有一个数字a[]，a的某子数组b[]如果能表示出a，则称b为好子数组。能表示出的意思即为，可以写出如下等式：对任意a[i]均有，a[i] = b[0]*t0 + b[1]*t1 + b[2]*t2 + b[3]*t3 + ... + b[m-1]*t(m-1).(b中有m个元素)。给定a，求它的元素数量最少的好子数组b。

　　　a[i] <= 10^7

解法：首先，从a中选出一些元素形成所求的b，其余元素构成数组b'。则一定有，b中所有元素的最大公约数一定是b'中所有元素的最大公约数的约数，否则无法表示。

　　　所以，为了方便，先将a中所有元素都除以最大公约数，然后在其中找到最少个数，使得找出的数最大公约数为1即可。

　　　其次，注意到，a[i]中每个元素含有质因子的个数最多为8个。(2*3*...23>10^7)

　　　所以，由于b中至少含有一个元素，所以对所有a[i]，枚举a[i]在b中的时候，b中最少含有多少个数，问题可以由集合dp得到解决。

tag:math, dp, good

  1 // BEGIN CUT HERE
  2 /*
  3 
  4 */
  5 // END CUT HERE
  6 #line 7 "BalanceScale.cpp"
  7 #include <cstdlib>
  8 #include <cctype>
  9 #include <cstring>
 10 #include <cstdio>
 11 #include <cmath>
 12 #include <algorithm>
 13 #include <vector>
 14 #include <iostream>
 15 #include <sstream>
 16 #include <set>
 17 #include <queue>
 18 #include <fstream>
 19 #include <numeric>
 20 #include <iomanip>
 21 #include <bitset>
 22 #include <list>
 23 #include <stdexcept>
 24 #include <functional>
 25 #include <string>
 26 #include <utility>
 27 #include <map>
 28 #include <ctime>
 29 #include <stack>
 30 
 31 using namespace std;
 32 
 33 #define clr(x) memset(x, 0, sizeof(x))
 34 #define clrs(x,y) memset(x, y, sizeof(x))
 35 #define pb push_back
 36 #define sz(v) ((int)(v).size())
 37 #define all(v) (v).begin(), (v).end()
 38 #define out(x) cout<<#x<<":"<<(x)<<endl
 39 #define tst(x) cout<<x<<" "
 40 #define tst1(a) cout<<#a<<endl
 41 #define CINBEQUICKER std::ios::sync_with_stdio(false)
 42 
 43 typedef long long int64;
 44 typedef vector<int> vi;
 45 typedef vector<string> vs;
 46 typedef vector<double> vd;
 47 typedef pair<int, int> pii;
 48 
 49 const double eps = 1e-8;
 50 const double PI = atan(1.0)*4;
 51 const int inf = 2139062143 / 2;
 52 
 53 inline int MyMod( int a , int b ) { return (a%b+b)%b;}
 54 
 55 int d[55][1<<9];
 56 vi divv;
 57 
 58 class BalanceScale
 59 {
 60     public:
 61         int gao(int sta, int x)
 62         {
 63             int m = sz(divv), ret = 0;
 64             for (int i = 0; i < m; ++ i) if (sta & (1<<i)){
 65                 if (x % divv[i] == 0) ret |= 1 << i;
 66             }
 67             return ret;
 68         }
 69         int takeWeights(vector <int> w){
 70             sort(all(w));
 71             int gd = w[0], n = sz(w);
 72             for (int i = 1; i < n; ++ i) gd = __gcd(gd, w[i]);
 73             for (int i = 0; i < n; ++ i){
 74                 w[i] /= gd;
 75                 if (w[i] == 1) return 1;
 76             }
 77             int ans = n;
 78             for (int i = 0; i < n; ++ i){
 79                 int tmp = w[i];
 80                 divv.clear();
 81                 for (int64 j = 2; j*j <= tmp; ++ j) if (tmp % j == 0){
 82                     divv.pb (j);
 83                     while (tmp % j == 0) tmp /= j;
 84                 }
 85                 if (tmp != 1) divv.pb (tmp);
 86 
 87                 int m = sz(divv);
 88                 clr (d); d[0][(1<<m)-1] = 1;
 89                 for (int j = 0; j < n; ++ j){
 90                     for (int k = 0; k <= j; ++ k)
 91                         for (int t = 0; t < (1<<m); ++ t) 
 92                             if (d[k][t]) d[k+1][gao(t, w[j])] = 1;
 93                 }
 94 
 95                 for (int j = 1; j < n; ++ j) if (d[j][0]) ans = min(ans, j+1);
 96             }
 97             return ans;
 98         }
 99         
100 // BEGIN CUT HERE
101     public:
102     void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); }
103     private:
104     template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
105     void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
106     //void test_case_0() { int Arr0[] = { 5, 4, 1, 8 }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 1; verify_case(0, Arg1, takeWeights(Arg0)); }
107     void test_case_0() { int Arr0[] = {3828825, 2159850, 4594590, 22, 1939392}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 1; verify_case(0, Arg1, takeWeights(Arg0)); }
108     void test_case_1() { int Arr0[] = { 2, 3, 8, 9 }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 2; verify_case(1, Arg1, takeWeights(Arg0)); }
109     void test_case_2() { int Arr0[] = { 60, 105, 490, 42 }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 4; verify_case(2, Arg1, takeWeights(Arg0)); }
110     void test_case_3() { int Arr0[] = { 15, 25, 9 }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 2; verify_case(3, Arg1, takeWeights(Arg0)); }
111 
112 // END CUT HERE
113 
114 };
115 //by plum rain
116 // BEGIN CUT HERE
117 int main()
118 {
119     //freopen( "a.out" , "w" , stdout );    
120     BalanceScale ___test;
121     ___test.run_test(-1);
122        return 0;
123 }
124 // END CUT HERE
View Code

 

　　　

 

posted @ 2014-01-29 15:36  Plumrain  阅读(310)  评论(0)  编辑  收藏  举报