标签:
题意:大概是求前n个数的最大公倍数。
解题思路:筛法+质因子分解,敲玩就去睡觉了,没想到没优化被cha了。。。
解题代码:
1 // BEGIN CUT HERE 2 /* 3 4 */ 5 // END CUT HERE 6 #line 7 "ThePermutationGame.cpp" 7 #include <cstdlib> 8 #include <cctype> 9 #include <cstring> 10 #include <cstdio> 11 #include <cmath> 12 #include <algorithm> 13 #include <vector> 14 #include <string> 15 #include <iostream> 16 #include <sstream> 17 #include <map> 18 #include <set> 19 #include <queue> 20 #include <stack> 21 #include <fstream> 22 #include <numeric> 23 #include <iomanip> 24 #include <bitset> 25 #include <list> 26 #include <stdexcept> 27 #include <functional> 28 #include <utility> 29 #include <ctime> 30 using namespace std; 31 32 #define PB push_back 33 #define MP make_pair 34 35 #define REP(i,n) for(i=0;i<(n);++i) 36 #define FOR(i,l,h) for(i=(l);i<=(h);++i) 37 #define FORD(i,h,l) for(i=(h);i>=(l);--i) 38 39 typedef vector<int> VI; 40 typedef vector<string> VS; 41 typedef vector<double> VD; 42 typedef long long LL; 43 typedef pair<int,int> PII; 44 #define maxn 1000005 45 #define M 1000000007 46 int hs[maxn]; 47 int a[maxn]; 48 int t = 0 ; 49 int solve() 50 { 51 for(int i = 2;i <= sqrt(100000);i ++) 52 { 53 if(hs[i] == 0) 54 { 55 int k = i * i ; 56 while(k <= 100000) 57 { 58 hs[k] = 1; 59 k = k + i ; 60 } 61 } 62 } 63 for(int i = 2;i <= 100000;i ++) 64 { 65 if(hs[i] == 0 ) 66 { 67 t ++; 68 a[t] = i ; 69 } 70 } 71 72 } 73 int ans[10000]; 74 int tmp[10000]; 75 LL P(LL x,LL y) 76 { 77 if(y == 0 ) 78 return 1; 79 LL ttt = P(x,y/2); 80 if(y%2 == 0 ) 81 return ttt*ttt % M; 82 else return ((ttt*ttt)%M)*x%M; 83 } 84 class ThePermutationGame 85 { 86 public: 87 int findMin(int N) 88 { 89 memset(hs,0,sizeof(hs)); 90 t = 0 ; 91 solve(); 92 printf("%d\n",t); 93 for(int i = 1;i <= N;i++) 94 { 95 memset(tmp,0,sizeof(tmp)); 96 int k = i ; 97 int tt = 1 ; 98 while(k!=1) 99 { 100 if(k %a[tt] == 0 ) 101 { 102 k = k/a[tt]; 103 tmp[tt] ++ ; 104 }else{ 105 tt ++ ; 106 } 107 } 108 for(int j = 1; j <= tt;j ++) 109 { 110 ans[j] = max(ans[j],tmp[j]); 111 } 112 } 113 LL sum = 1; 114 for(int j = 1;j <= t; j ++) 115 { 116 sum = sum * P(a[j],ans[j]) % M; 117 } 118 return sum; 119 } 120 121 // BEGIN CUT HERE 122 public: 123 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(); } 124 private: 125 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(); } 126 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; } } 127 void test_case_0() { int Arg0 = 2; int Arg1 = 2; verify_case(0, Arg1, findMin(Arg0)); } 128 void test_case_1() { int Arg0 = 3; int Arg1 = 6; verify_case(1, Arg1, findMin(Arg0)); } 129 void test_case_2() { int Arg0 = 11; int Arg1 = 504; verify_case(2, Arg1, findMin(Arg0)); } 130 void test_case_3() { int Arg0 = 102; int Arg1 = 841777601; verify_case(3, Arg1, findMin(Arg0)); } 131 void test_case_4() { int Arg0 = 9999; int Arg1 = 804862568; verify_case(4, Arg1, findMin(Arg0)); } 132 133 // END CUT HERE 134 135 }; 136 137 // BEGIN CUT HERE 138 int main() 139 { 140 ThePermutationGame ___test; 141 ___test.run_test(-1); 142 return 0; 143 } 144 // END CUT HERE
标签:
原文地址:http://www.cnblogs.com/zyue/p/4324989.html