SRM 609(1-250pt, 1-500pt)
嗯。。。。还是应该坚持写题解的好习惯啊。。。
DIV1 250pt
这难度是回到srm 300+的250了嘛。。。略
1 // BEGIN CUT HERE 2 /* 3 * Author: plum rain 4 * score : 5 */ 6 /* 7 8 */ 9 // END CUT HERE 10 #line 11 "MagicalStringDiv1.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 MagicalStringDiv1 59 { 60 public: 61 int getLongest(string s){ 62 int t1 = 0, t2 = sz(s)-1, num = 0; 63 while (t1 <= t2){ 64 while (t1 < sz(s) && s[t1] != '>') ++ t1; 65 while (t2 >= 0 && s[t2] != '<') -- t2; 66 if (t1 <= t2) -- t2, ++ t1, num += 2; 67 } 68 return num; 69 } 70 71 // BEGIN CUT HERE 72 public: 73 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(); } 74 private: 75 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(); } 76 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; } } 77 void test_case_0() { string Arg0 = "<"; int Arg1 = 4; verify_case(0, Arg1, getLongest(Arg0)); } 78 void test_case_1() { string Arg0 = ">>><<<"; int Arg1 = 6; verify_case(1, Arg1, getLongest(Arg0)); } 79 void test_case_2() { string Arg0 = "<<<>>>"; int Arg1 = 0; verify_case(2, Arg1, getLongest(Arg0)); } 80 void test_case_3() { string Arg0 = "<<<<><>>><>>><>><>><>>><<<<>><>>>>><<>>>>><><<<<>>"; int Arg1 = 24; verify_case(3, Arg1, getLongest(Arg0)); } 81 82 // END CUT HERE 83 84 }; 85 86 // BEGIN CUT HERE 87 int main() 88 { 89 // freopen( "a.out" , "w" , stdout ); 90 MagicalStringDiv1 ___test; 91 ___test.run_test(-1); 92 return 0; 93 } 94 // END CUT HERE
DIV1 500pt
题意:有k种颜色的球,每种颜色有an[i]个(an[i] > 0)。有两种背包,每种背包都有无限多个,第一种背包里面装的所有球颜色必须相同,第二种背包里面装的球颜色不能有相同的。无论哪一种背包,最多都只能装k个。问,最少用多少个背包能装下所有球。
an[i] <= 10^9, k <= 10^5
解法:嗯。。。首先,如果不要求每个背包最多装k个,那就很简单了。直接用an[i]枚举有多少个第二种背包,然后第一种背包的数量就是n - i - 1。
然后,加上每个背包只能装k个的限制条件,就是首先对k个背包预处理一遍,an[i] %= k。后面一步是yy出来的。。。感觉是这样,不知道怎么证。。。
tag:think
1 // BEGIN CUT HERE 2 /* 3 * Author: plum rain 4 * score : 5 */ 6 /* 7 8 */ 9 // END CUT HERE 10 #line 11 "PackingBallsDiv1.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 int an[100005]; 59 vi yu; 60 61 class PackingBallsDiv1 62 { 63 public: 64 int minPacks(int n, int a, int b, int c, int mod){ 65 an[0] = a; 66 for (int i = 1; i < n; ++ i) an[i] = ((int64)an[i-1] * b + c) % mod + 1; 67 68 int64 num = 0; 69 for (int i = 0; i < n; ++ i){ 70 num += an[i] / n; 71 an[i] %= n; 72 } 73 sort (an, an+n); 74 75 //if (n == 3) for (int i = 0; i < n; ++ i) tst (an[i]); 76 //cout << endl; 77 // 78 int64 cnt = n; 79 for (int i = 0; i < n; ++ i){ 80 int64 tmp = an[i] + n - i - 1; 81 cnt = cnt > tmp ? tmp : cnt; 82 } 83 return (int)(cnt + num); 84 } 85 86 // BEGIN CUT HERE 87 public: 88 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(); } 89 private: 90 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(); } 91 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; } } 92 void test_case_0() { int Arg0 = 3; int Arg1 = 4; int Arg2 = 2; int Arg3 = 5; int Arg4 = 6; int Arg5 = 4; verify_case(0, Arg5, minPacks(Arg0, Arg1, Arg2, Arg3, Arg4)); } 93 void test_case_1() { int Arg0 = 1; int Arg1 = 58; int Arg2 = 23; int Arg3 = 39; int Arg4 = 93; int Arg5 = 58; verify_case(1, Arg5, minPacks(Arg0, Arg1, Arg2, Arg3, Arg4)); } 94 void test_case_2() { int Arg0 = 23; int Arg1 = 10988; int Arg2 = 5573; int Arg3 = 4384; int Arg4 = 100007; int Arg5 = 47743; verify_case(2, Arg5, minPacks(Arg0, Arg1, Arg2, Arg3, Arg4)); } 95 void test_case_3() { int Arg0 = 100000; int Arg1 = 123456789; int Arg2 = 234567890; int Arg3 = 345678901; int Arg4 = 1000000000; int Arg5 = 331988732; verify_case(3, Arg5, minPacks(Arg0, Arg1, Arg2, Arg3, Arg4)); } 96 97 // END CUT HERE 98 99 }; 100 101 // BEGIN CUT HERE 102 int main() 103 { 104 //freopen( "a.out" , "w" , stdout ); 105 PackingBallsDiv1 ___test; 106 ___test.run_test(-1); 107 return 0; 108 } 109 // END CUT HERE
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现在的你,在干什么呢?
你是不是还记得,你说你想成为岩哥那样的人。
现在的你,在干什么呢?
你是不是还记得,你说你想成为岩哥那样的人。