SRM 502(2-1000pt)
题意:在0~(n-1)中选择k个数,使得他们的和为n的倍数的选择方案有多少种。(n <= 1000, k <= 47)
解法:裸dp。d[i][j][k’]表示在前i个数中(0~i-1),选择k‘个数使得其和mod n的余数为j的选择方案的种数。时间、空间复杂度均为O(n^2*k),时间复杂度能接受,空间不能,用滚动数组优化即可。
tag:dp, 水题
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1 // BEGIN CUT HERE 2 /* 3 * Author: plum rain 4 * score : 5 */ 6 /* 7 8 */ 9 // END CUT HERE 10 #line 11 "TheCowDivTwo.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 CLR(x) memset(x, 0, sizeof(x)) 38 #define PB push_back 39 #define SZ(v) ((int)(v).size()) 40 #define zero(x) (((x)>0?(x):-(x))<eps) 41 #define out(x) cout<<#x<<":"<<(x)<<endl 42 #define tst(a) cout<<#a<<endl 43 #define CINBEQUICKER std::ios::sync_with_stdio(false) 44 45 typedef vector<int> VI; 46 typedef vector<string> VS; 47 typedef vector<double> VD; 48 typedef long long int64; 49 50 const double eps = 1e-8; 51 const double PI = atan(1.0)*4; 52 const int maxint = 2139062143; 53 const int mod = 1000000007; 54 55 int d[2][1005][50]; 56 57 class TheCowDivTwo 58 { 59 public: 60 int find(int n, int K){ 61 CLR (d); 62 d[0][0][0] = 1; 63 int cur = 1; 64 for (int i = 1; i <= n; ++ i){ 65 for (int j = 0; j < n; ++ j){ 66 int tmp = j-(i-1)<0 ? n+j-(i-1) : j-(i-1); 67 for (int k = 0; k <= K; ++ k){ 68 d[cur][j][k] = d[cur^1][j][k]; 69 if (k){ 70 d[cur][j][k] = (d[cur][j][k] + d[cur^1][tmp][k-1]) % mod; 71 } 72 } 73 } 74 cur ^= 1; 75 } 76 77 return d[cur^1][0][K]; 78 } 79 80 // BEGIN CUT HERE 81 public: 82 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(); } 83 private: 84 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(); } 85 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; } } 86 void test_case_0() { int Arg0 = 7; int Arg1 = 4; int Arg2 = 5; verify_case(0, Arg2, find(Arg0, Arg1)); } 87 void test_case_1() { int Arg0 = 1; int Arg1 = 1; int Arg2 = 1; verify_case(1, Arg2, find(Arg0, Arg1)); } 88 void test_case_2() { int Arg0 = 58; int Arg1 = 4; int Arg2 = 7322; verify_case(2, Arg2, find(Arg0, Arg1)); } 89 void test_case_3() { int Arg0 = 502; int Arg1 = 7; int Arg2 = 704466492; verify_case(3, Arg2, find(Arg0, Arg1)); } 90 void test_case_4() { int Arg0 = 1000; int Arg1 = 47; int Arg2 = 219736903; verify_case(4, Arg2, find(Arg0, Arg1)); } 91 92 // END CUT HERE 93 94 }; 95 96 // BEGIN CUT HERE 97 int main() 98 { 99 // freopen( "a.out" , "w" , stdout ); 100 TheCowDivTwo ___test; 101 ___test.run_test(-1); 102 return 0; 103 } 104 // 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。