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