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

 // BEGIN CUT HERE
 /*

 */
 // END CUT HERE
 #line 7 "Underprimes.cpp"
 #include <cstdlib>
 #include <cctype>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <iostream>
 #include <sstream>
 #include <set>
 #include <queue>
 #include <fstream>
 #include <numeric>
 #include <iomanip>
 #include <bitset>
 #include <list>
 #include <stdexcept>
 #include <functional>
 #include <string>
 #include <utility>
 #include <map>
 #include <ctime>
 #include <stack>

 using namespace std;

 #define clr0(x) memset(x, 0, sizeof(x))
 #define clr1(x) memset(x, -1, sizeof(x))
 #define pb push_back
 #define mp make_pair
 #define sz(v) ((int)(v).size())
 #define out(x) cout<<#x<<":"<<(x)<<endl
 #define tst(a) cout<<#a<<endl
 #define CINBEQUICKER std::ios::sync_with_stdio(false)

 typedef vector<int> VI;
 typedef vector<string> VS;
 typedef vector<double> VD;
 typedef long long int64;

 const double eps = 1e-;
 const double PI = atan(1.0)*;
 const int inf =  / ;

 inline int MyMod( int a , int b ) { return (a%b+b)%b;}
 int temp[] = {,,,,,,,};

 class Underprimes
 {
     public:
         bool gao(int x)
         {
             int cnt = ;
             for (int i = ; i*i <= x; ++ i) if (!(x % i)){
                 while (!(x % i)) x /= i, ++ cnt;
             }
             if (x != ) ++ cnt;
             for (int i = ; i < ; ++ i) if (cnt == temp[i]) return ;
             return ;
         }
         int howMany(int A, int B){
             int ret = ;
             for (int i = B; i >= A; -- i)
                 if (gao(i)) ++ ret;
             return ret;
         }

 // BEGIN CUT HERE
     public:
     void run_test(int Case) { if ((Case == -) || (Case == )) test_case_0(); if ((Case == -) || (Case == )) test_case_1(); if ((Case == -) || (Case == )) test_case_2(); if ((Case == -) || (Case == )) test_case_3(); }
     private:
     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(); }
     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; } }
     void test_case_0() { int Arg0 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, howMany(Arg0, Arg1)); }
     void test_case_1() { int Arg0 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, howMany(Arg0, Arg1)); }
     void test_case_2() { int Arg0 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, howMany(Arg0, Arg1)); }
     void test_case_3() { int Arg0 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, howMany(Arg0, Arg1)); }

 // END CUT HERE

 };
 //by plum rain
 // BEGIN CUT HERE
 int main()
 {
     //freopen( "a.out" , "w" , stdout );
     Underprimes ___test;
     ___test.run_test(-);
        return ;
 }
 // END CUT HERE

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

 // BEGIN CUT HERE
 /*
  * Author:  plum rain
  * score :
  */
 /*

  */
 // END CUT HERE
 #line 11 "BedroomFloor.cpp"
 #include <sstream>
 #include <stdexcept>
 #include <functional>
 #include <iomanip>
 #include <numeric>
 #include <fstream>
 #include <cctype>
 #include <iostream>
 #include <cstdio>
 #include <vector>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <cstdlib>
 #include <set>
 #include <queue>
 #include <bitset>
 #include <list>
 #include <string>
 #include <utility>
 #include <map>
 #include <ctime>
 #include <stack>

 using namespace std;

 #define clr0(x) memset(x, 0, sizeof(x))
 #define clr1(x) memset(x, -1, sizeof(x))
 #define pb push_back
 #define sz(v) ((int)(v).size())
 #define all(t) t.begin(),t.end()
 #define zero(x) (((x)>0?(x):-(x))<eps)
 #define out(x) cout<<#x<<":"<<(x)<<endl
 #define tst(a) cout<<a<<" "
 #define tst1(a) cout<<#a<<endl
 #define CINBEQUICKER std::ios::sync_with_stdio(false)

 typedef vector<int> vi;
 typedef vector<string> vs;
 typedef vector<double> vd;
 typedef pair<int, int> pii;
 typedef long long int64;

 const double eps = 1e-;
 const double PI = atan(1.0)*;
 const int inf =  / ;

 class BedroomFloor
 {
     public:
         int64 num[];

         int64 min(int64 a, int64 b)
         {
             return a > b ? b : a;
         }

         int64 gao(int64 sam, int64 s, int64 e, int64 len)
         {
             int64 tim =  e/ - s/ - ;
             if (tim < ){
                 if (sam) num[len] += e - s + ;
                 else num[e-s+] += len;
                 return ;
             }

             int64 ret = ;
             if (len == ) ret = tim * ;
             else{
                 num[len] += (tim + (sam^)) /  * ;
                 ret = (tim + sam) /  * len;
             }

             int64 up = (s/ + ) * ;
             if (!sam) num[up-s] += len;
             else num[len] += up - s;

             sam ^= (tim + ) & ;
             int64 dn = e /  *  - ;
             if (!sam) num[e-dn] += len;
             else num[len] += e - dn;

             return ret;
         }

         long long numberOfSticks(int x1, int y1, int x2, int y2){
             clr0 (num);
             int64 ret = ;
             for (int i = x1; i < x2;){
                 ret += gao(((i/) & ) == ((y1/) & ), y1, y2-, min( - (i%), x2-i));
                 i = i /  *  + ;
             }

             //for (int i = 1; i < 6; ++ i)
                 //tst(i), out (num[i]);
 //
             ret += num[];
             if (num[] && num[]){
                 int64 tmp = min(num[], num[]);
                 ret += tmp;
                 num[] -= tmp; num[] -= tmp;
             }
             ret += num[];

             if (num[] && num[]){
                 int64 tmp = min(num[], num[]);
                 ret += tmp;
                 num[] -= tmp; num[] -= tmp;
             }
             if (num[] && num[]){
                 int64 tmp = min(num[]/, num[]);
                 ret += tmp;
                 num[] -= *tmp; num[] -= tmp;
             }
             ret += num[];
             num[] -= num[] * ;

             if (num[] && num[] > ){
                 int64 tmp = min(num[]/, num[]);
                 ret += tmp;
                 num[] -= tmp*; num[] -= tmp;
             }
             if (num[] > ) ret += num[]& ? num[]/+ : num[]/;
             else if (num[] == ) ret += , num[] -= ;

             if (num[] > ) ret += num[]% ? num[]/+ : num[]/;
             return ret;
         }

 // BEGIN CUT HERE
     public:
     void run_test(int Case) { if ((Case == -) || (Case == )) test_case_0(); if ((Case == -) || (Case == )) test_case_1(); if ((Case == -) || (Case == )) test_case_2(); if ((Case == -) || (Case == )) test_case_3(); if ((Case == -) || (Case == )) test_case_4(); }
     //void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_1();}
     private:
     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(); }
     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 << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
     void test_case_0() { int Arg0 = ; int Arg1 = ; int Arg2 = ; int Arg3 = ; long long Arg4 = 5LL; verify_case(, Arg4, numberOfSticks(Arg0, Arg1, Arg2, Arg3)); }
     void test_case_1() { int Arg0 = ; int Arg1 = ; int Arg2 = ; int Arg3 = ; long long Arg4 = 5LL; verify_case(, Arg4, numberOfSticks(Arg0, Arg1, Arg2, Arg3)); }
     void test_case_2() { int Arg0 = ; int Arg1 = ; int Arg2 = ; int Arg3 = ; long long Arg4 = 12LL; verify_case(, Arg4, numberOfSticks(Arg0, Arg1, Arg2, Arg3)); }
     void test_case_3() { int Arg0 = ; int Arg1 = ; int Arg2 = ; int Arg3 = ; long long Arg4 = 27LL; verify_case(, Arg4, numberOfSticks(Arg0, Arg1, Arg2, Arg3)); }
     void test_case_4() { int Arg0 = ; int Arg1 = ; int Arg2 = ; int Arg3 = ; long long Arg4 = 200000000000LL; verify_case(, Arg4, numberOfSticks(Arg0, Arg1, Arg2, Arg3)); }

 // END CUT HERE

 };

 // BEGIN CUT HERE
 int main()
 {
 //    freopen( "a.out" , "w" , stdout );
     BedroomFloor ___test;
     ___test.run_test(-);
        return ;
 }
 // END CUT HERE