SRM 399(1-250pt)
DIV1 250pt
题意:给出一个size不超过50的数组a和整数n,求x,y,z使得|n - x*y*z|最小,且x,y,z均不再数组a中。若有多组xyz使得|n-x*y*z|最小,输出字典序最小的。a中的元素和n <= 1000。
解法:我直接写了一个暴力扫的方法,x*y*z上限设成了10000....发现如果a中含有1,2,3....50,就会WA,因为这样的话x*y*z的需要达到51*51*51。后来又写了一个方法也错了.....没做出来看了题解,发现它是div2的1000突然就感觉到安慰了.......
要观察到一个东西,就是x,y,z都不可能超过n + 51。然后暴力就可以了,加一个break语句,当x*y*z > 200000时。
想了一下,也就是说,其实可以不用观察到那个结论,只需要对xyz写3个for语句从1到1010,然后在x*y*z > 200000时break,这道题就能过了....
tag:think
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 // BEGIN CUT HERE 2 /* 3 * Author: plum rain 4 * score : 5 */ 6 /* 7 8 */ 9 // END CUT HERE 10 #line 11 "AvoidingProduct.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 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<<endl 45 #define CINBEQUICKER std::ios::sync_with_stdio(false) 46 47 typedef vector<int> VI; 48 typedef vector<string> VS; 49 typedef vector<double> VD; 50 typedef pair<int, int> pii; 51 typedef long long int64; 52 53 const double eps = 1e-8; 54 const double PI = atan(1.0)*4; 55 const int maxint = 2139062143; 56 const int N = 200000; 57 58 bool v[1100]; 59 60 class AvoidingProduct 61 { 62 public: 63 vector <int> getTriple(vector <int> a, int n){ 64 int sz = a.size(); 65 CLR (v); 66 for (int i = 0; i < sz; ++ i) v[a[i]] = 1; 67 68 int num = maxint; 69 VI cur, tmp; 70 for (int i = 1; i <= 1010; ++ i) if (!v[i]){ 71 if (i*i*i > N) break; 72 for (int j = i; j <= 1010; ++ j) if (!v[j]){ 73 if (i*j*j > N) break; 74 for (int k = j; k <= 1010; ++ k) if (!v[k]){ 75 int mul = i * j * k; 76 int t_num = abs(n - mul); 77 if (num == t_num){ 78 tmp.clear(); 79 tmp.PB (i); tmp.PB (j); tmp.PB (k); 80 } 81 if (t_num < num || (t_num == num && tmp < cur)){ 82 num = t_num; 83 cur.clear(); 84 cur.PB (i); cur.PB (j); cur.PB (k); 85 } 86 if (mul > N) break; 87 } 88 } 89 } 90 return cur; 91 } 92 93 // BEGIN CUT HERE 94 public: 95 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(); if ((Case == -1) || (Case == 5)) test_case_5(); } 96 private: 97 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(); } 98 void verify_case(int Case, const vector <int> &Expected, const vector <int> &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: " << print_array(Expected) << endl; cerr << "\tReceived: " << print_array(Received) << endl; } } 99 void test_case_0() { int Arr0[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 1000; int Arr2[] = {51, 51, 51 }; vector <int> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[0]))); verify_case(0, Arg2, getTriple(Arg0, Arg1)); } 100 void test_case_1() { int Arr0[] = {1}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 10; int Arr2[] = {2, 2, 2 }; vector <int> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[0]))); verify_case(1, Arg2, getTriple(Arg0, Arg1)); } 101 void test_case_2() { int Arr0[] = {1,2}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 10; int Arr2[] = {3, 3, 3 }; vector <int> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[0]))); verify_case(2, Arg2, getTriple(Arg0, Arg1)); } 102 void test_case_3() { int Arr0[] = {1,3}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 12; int Arr2[] = {2, 2, 2 }; vector <int> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[0]))); verify_case(3, Arg2, getTriple(Arg0, Arg1)); } 103 void test_case_4() { int Arr0[] = {1,3}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 13; int Arr2[] = {2, 2, 4 }; vector <int> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[0]))); verify_case(4, Arg2, getTriple(Arg0, Arg1)); } 104 void test_case_5() { int Arr0[] = {1,15}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 90; int Arr2[] = {2, 5, 9 }; vector <int> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[0]))); verify_case(5, Arg2, getTriple(Arg0, Arg1)); } 105 106 // END CUT HERE 107 108 }; 109 110 // BEGIN CUT HERE 111 int main() 112 { 113 // freopen( "a.out" , "w" , stdout ); 114 AvoidingProduct ___test; 115 ___test.run_test(-1); 116 return 0; 117 } 118 // 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。