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-8;
const double PI = atan(1.0)*4;
const int inf = 2139062143 / 2;

inline int MyMod( int a , int b ) { return (a%b+b)%b;}
int temp[] = {2,3,5,7,11,13,17,19};

class Underprimes
{
    public:
        bool gao(int x)
        {
            int cnt = 0;
            for (int i = 2; i*i <= x; ++ i) if (!(x % i)){
                while (!(x % i)) x /= i, ++ cnt;
            }
            if (x != 1) ++ cnt;
            for (int i = 0; i < 8; ++ i) if (cnt == temp[i]) return 1;
            return 0;
        }
        int howMany(int A, int B){
            int ret = 0;
            for (int i = B; i >= A; -- i)
                if (gao(i)) ++ ret;
            return ret;
        }
        
// BEGIN CUT HERE
    public:
    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(); }
    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 << "	Expected: "" << Expected << '"' << endl; cerr << "	Received: "" << Received << '"' << endl; } }
    void test_case_0() { int Arg0 = 2; int Arg1 = 10; int Arg2 = 5; verify_case(0, Arg2, howMany(Arg0, Arg1)); }
    void test_case_1() { int Arg0 = 100; int Arg1 = 105; int Arg2 = 2; verify_case(1, Arg2, howMany(Arg0, Arg1)); }
    void test_case_2() { int Arg0 = 17; int Arg1 = 17; int Arg2 = 0; verify_case(2, Arg2, howMany(Arg0, Arg1)); }
    void test_case_3() { int Arg0 = 123; int Arg1 = 456; int Arg2 = 217; verify_case(3, 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(-1);
       return 0;
}
// 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-8;
const double PI = atan(1.0)*4;
const int inf = 2139062143 / 2;

class BedroomFloor
{
    public:
        int64 num[10];

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

        int64 gao(int64 sam, int64 s, int64 e, int64 len)
        {
            int64 tim =  e/5 - s/5 - 1;
            if (tim < 0){
                if (sam) num[len] += e - s + 1;
                else num[e-s+1] += len;
                return 0;
            }

            int64 ret = 0;
            if (len == 5) ret = tim * 5;
            else{
                num[len] += (tim + (sam^1)) / 2 * 5;
                ret = (tim + sam) / 2 * len; 
            }

            int64 up = (s/5 + 1) * 5;
            if (!sam) num[up-s] += len;
            else num[len] += up - s;

            sam ^= (tim + 1) & 1;
            int64 dn = e / 5 * 5 - 1;
            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 = 0;
            for (int i = x1; i < x2;){
                ret += gao(((i/5) & 1) == ((y1/5) & 1), y1, y2-1, min(5 - (i%5), x2-i));
                i = i / 5 * 5 + 5;
            }

            //for (int i = 1; i < 6; ++ i)
                //tst(i), out (num[i]);
//
            ret += num[5];
            if (num[4] && num[1]){
                int64 tmp = min(num[4], num[1]);
                ret += tmp;
                num[4] -= tmp; num[1] -= tmp;
            }
            ret += num[4];

            if (num[3] && num[2]){
                int64 tmp = min(num[3], num[2]);
                ret += tmp;
                num[3] -= tmp; num[2] -= tmp;
            }
            if (num[3] && num[1]){
                int64 tmp = min(num[1]/2, num[3]);
                ret += tmp;
                num[1] -= 2*tmp; num[3] -= tmp;
            }
            ret += num[3];
            num[1] -= num[3] * 2;

            if (num[2] && num[1] > 0){
                int64 tmp = min(num[2]/2, num[1]);
                ret += tmp;
                num[2] -= tmp*2; num[1] -= tmp;
            }
            if (num[2] > 1) ret += num[2]&1 ? num[2]/2+1 : num[2]/2;
            else if (num[2] == 1) ret += 1, num[1] -= 3;

            if (num[1] > 0) ret += num[1]%5 ? num[1]/5+1 : num[1]/5;
            return ret;
        }
        
// BEGIN CUT HERE
    public:
    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(); }
    //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 << "	Expected: "" << Expected << '"' << endl; cerr << "	Received: "" << Received << '"' << endl; } }
    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)); }
    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)); }
    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)); }
    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)); }
    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)); }

// END CUT HERE

};

// BEGIN CUT HERE
int main()
{
//    freopen( "a.out" , "w" , stdout );    
    BedroomFloor ___test;
    ___test.run_test(-1);
       return 0;
}
// END CUT HERE