SRM 442(1-250pt, 1-500pt)

DIV1 250pt

题意：将一个数表示成质因子相乘的形式，若乘式所含数字的个数为质数，则称A为underprime。比如12 = 2*2*3，则含3个数字，是underprime。求A, B之间underprime的个数。A, B <= 10^5。

解法：暴力枚举A,B之间所有数，求出其乘式所含数字的个数， 判断是不是质数。

tag:brute-force

 1 // BEGIN CUT HERE
 2 /*
 3 
 4 */
 5 // END CUT HERE
 6 #line 7 "Underprimes.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(a) cout<<#a<<endl
40 #define CINBEQUICKER std::ios::sync_with_stdio(false)
41 
42 typedef vector<int> VI;
43 typedef vector<string> VS;
44 typedef vector<double> VD;
45 typedef long long int64;
46 
47 const double eps = 1e-8;
48 const double PI = atan(1.0)*4;
49 const int inf = 2139062143 / 2;
50 
51 inline int MyMod( int a , int b ) { return (a%b+b)%b;}
52 int temp[] = {2,3,5,7,11,13,17,19};
53 
54 class Underprimes
55 {
56     public:
57         bool gao(int x)
58         {
59             int cnt = 0;
60             for (int i = 2; i*i <= x; ++ i) if (!(x % i)){
61                 while (!(x % i)) x /= i, ++ cnt;
62             }
63             if (x != 1) ++ cnt;
64             for (int i = 0; i < 8; ++ i) if (cnt == temp[i]) return 1;
65             return 0;
66         }
67         int howMany(int A, int B){
68             int ret = 0;
69             for (int i = B; i >= A; -- i)
70                 if (gao(i)) ++ ret;
71             return ret;
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(); }
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 << "	Expected: "" << Expected << '"' << endl; cerr << "	Received: "" << Received << '"' << endl; } }
80     void test_case_0() { int Arg0 = 2; int Arg1 = 10; int Arg2 = 5; verify_case(0, Arg2, howMany(Arg0, Arg1)); }
81     void test_case_1() { int Arg0 = 100; int Arg1 = 105; int Arg2 = 2; verify_case(1, Arg2, howMany(Arg0, Arg1)); }
82     void test_case_2() { int Arg0 = 17; int Arg1 = 17; int Arg2 = 0; verify_case(2, Arg2, howMany(Arg0, Arg1)); }
83     void test_case_3() { int Arg0 = 123; int Arg1 = 456; int Arg2 = 217; verify_case(3, Arg2, howMany(Arg0, Arg1)); }
84 
85 // END CUT HERE
86 
87 };
88 //by plum rain
89 // BEGIN CUT HERE
90 int main()
91 {
92     //freopen( "a.out" , "w" , stdout );    
93     Underprimes ___test;
94     ___test.run_test(-1);
95        return 0;
96 }
97 // END CUT HERE
View Code

DIV1 550pt

题意：用1*5的瓷砖将房间铺成如下图形状，现在给一个矩形的区域，重新铺里面的瓷砖。买一个1*5的瓷砖，可以将其切割成几小块使用，但不能将几小块合并成一个1*5的使用。问要重新铺所给区域瓷砖，最少买多少块1*5的瓷砖。

解法：分两步，第一步求出，给定区域内，含有1*5，1*4，1*3，1*2，1*1各多少块。这一步只能暴力求，不同的写法上可能有编码复杂度的差异，我觉得我的还行，就是效率不太高。。

　　　第二步，求出各种块的数量之后，求最少买多少块瓷砖。贪心思想，方法是从大往小扫，如果有1*4的，每块都配一个1*1的只要有，如果有1*3的尽量配1*2，不行就配1*1。。。。就是从大往小扫，匹配能匹配到的最大的瓷砖。这样的匹配方法在SRM 598 DIV1的250pt里面有用到。

tag:brute-force, greedy

  1 // BEGIN CUT HERE
  2 /*
  3  * Author:  plum rain
  4  * score :
  5  */
  6 /*
  7 
  8  */
  9 // END CUT HERE
 10 #line 11 "BedroomFloor.cpp"
 11 #include <sstream>
 12 #include <stdexcept>
 13 #include <functional>
 14 #include <iomanip>
 15 #include <numeric>
 16 #include <fstream>
 17 #include <cctype>
 18 #include <iostream>
 19 #include <cstdio>
 20 #include <vector>
 21 #include <cstring>
 22 #include <cmath>
 23 #include <algorithm>
 24 #include <cstdlib>
 25 #include <set>
 26 #include <queue>
 27 #include <bitset>
 28 #include <list>
 29 #include <string>
 30 #include <utility>
 31 #include <map>
 32 #include <ctime>
 33 #include <stack>
 34 
 35 using namespace std;
 36 
 37 #define clr0(x) memset(x, 0, sizeof(x))
 38 #define clr1(x) memset(x, -1, sizeof(x))
 39 #define pb push_back
 40 #define sz(v) ((int)(v).size())
 41 #define all(t) t.begin(),t.end()
 42 #define zero(x) (((x)>0?(x):-(x))<eps)
 43 #define out(x) cout<<#x<<":"<<(x)<<endl
 44 #define tst(a) cout<<a<<" "
 45 #define tst1(a) cout<<#a<<endl
 46 #define CINBEQUICKER std::ios::sync_with_stdio(false)
 47 
 48 typedef vector<int> vi;
 49 typedef vector<string> vs;
 50 typedef vector<double> vd;
 51 typedef pair<int, int> pii;
 52 typedef long long int64;
 53 
 54 const double eps = 1e-8;
 55 const double PI = atan(1.0)*4;
 56 const int inf = 2139062143 / 2;
 57 
 58 class BedroomFloor
 59 {
 60     public:
 61         int64 num[10];
 62 
 63         int64 min(int64 a, int64 b)
 64         {
 65             return a > b ? b : a;
 66         }
 67 
 68         int64 gao(int64 sam, int64 s, int64 e, int64 len)
 69         {
 70             int64 tim =  e/5 - s/5 - 1;
 71             if (tim < 0){
 72                 if (sam) num[len] += e - s + 1;
 73                 else num[e-s+1] += len;
 74                 return 0;
 75             }
 76 
 77             int64 ret = 0;
 78             if (len == 5) ret = tim * 5;
 79             else{
 80                 num[len] += (tim + (sam^1)) / 2 * 5;
 81                 ret = (tim + sam) / 2 * len; 
 82             }
 83 
 84             int64 up = (s/5 + 1) * 5;
 85             if (!sam) num[up-s] += len;
 86             else num[len] += up - s;
 87 
 88             sam ^= (tim + 1) & 1;
 89             int64 dn = e / 5 * 5 - 1;
 90             if (!sam) num[e-dn] += len;
 91             else num[len] += e - dn;
 92 
 93             return ret;
 94         }
 95 
 96         long long numberOfSticks(int x1, int y1, int x2, int y2){
 97             clr0 (num);
 98             int64 ret = 0;
 99             for (int i = x1; i < x2;){
100                 ret += gao(((i/5) & 1) == ((y1/5) & 1), y1, y2-1, min(5 - (i%5), x2-i));
101                 i = i / 5 * 5 + 5;
102             }
103 
104             //for (int i = 1; i < 6; ++ i)
105                 //tst(i), out (num[i]);
106 //
107             ret += num[5];
108             if (num[4] && num[1]){
109                 int64 tmp = min(num[4], num[1]);
110                 ret += tmp;
111                 num[4] -= tmp; num[1] -= tmp;
112             }
113             ret += num[4];
114 
115             if (num[3] && num[2]){
116                 int64 tmp = min(num[3], num[2]);
117                 ret += tmp;
118                 num[3] -= tmp; num[2] -= tmp;
119             }
120             if (num[3] && num[1]){
121                 int64 tmp = min(num[1]/2, num[3]);
122                 ret += tmp;
123                 num[1] -= 2*tmp; num[3] -= tmp;
124             }
125             ret += num[3];
126             num[1] -= num[3] * 2;
127 
128             if (num[2] && num[1] > 0){
129                 int64 tmp = min(num[2]/2, num[1]);
130                 ret += tmp;
131                 num[2] -= tmp*2; num[1] -= tmp;
132             }
133             if (num[2] > 1) ret += num[2]&1 ? num[2]/2+1 : num[2]/2;
134             else if (num[2] == 1) ret += 1, num[1] -= 3;
135 
136             if (num[1] > 0) ret += num[1]%5 ? num[1]/5+1 : num[1]/5;
137             return ret;
138         }
139         
140 // BEGIN CUT HERE
141     public:
142     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(); }
143     //void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_1();}
144     private:
145     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(); }
146     void verify_case(int Case, const long long &Expected, const long long &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "	Expected: "" << Expected << '"' << endl; cerr << "	Received: "" << Received << '"' << endl; } }
147     void test_case_0() { int Arg0 = 0; int Arg1 = 0; int Arg2 = 5; int Arg3 = 5; long long Arg4 = 5LL; verify_case(0, Arg4, numberOfSticks(Arg0, Arg1, Arg2, Arg3)); }
148     void test_case_1() { int Arg0 = 0; int Arg1 = 0; int Arg2 = 10; int Arg3 = 2; long long Arg4 = 5LL; verify_case(1, Arg4, numberOfSticks(Arg0, Arg1, Arg2, Arg3)); }
149     void test_case_2() { int Arg0 = 2; int Arg1 = 2; int Arg2 = 8; int Arg3 = 8; long long Arg4 = 12LL; verify_case(2, Arg4, numberOfSticks(Arg0, Arg1, Arg2, Arg3)); }
150     void test_case_3() { int Arg0 = 8; int Arg1 = 5; int Arg2 = 20; int Arg3 = 16; long long Arg4 = 27LL; verify_case(3, Arg4, numberOfSticks(Arg0, Arg1, Arg2, Arg3)); }
151     void test_case_4() { int Arg0 = 0; int Arg1 = 0; int Arg2 = 1000000; int Arg3 = 1000000; long long Arg4 = 200000000000LL; verify_case(4, Arg4, numberOfSticks(Arg0, Arg1, Arg2, Arg3)); }
152 
153 // END CUT HERE
154 
155 };
156 
157 // BEGIN CUT HERE
158 int main()
159 {
160 //    freopen( "a.out" , "w" , stdout );    
161     BedroomFloor ___test;
162     ___test.run_test(-1);
163        return 0;
164 }
165 // END CUT HERE
View Code
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原文地址：https://www.cnblogs.com/plumrain/p/srm_442.html