题意：给你一张长W，宽H 的纸，问你能不能者成给定的大小， 每一次折纸只能折成整数大小。
解题思路：递推+枚举 枚举给定大小的长和宽，然后套进 W，H中求最小值 ， 折纸策略最优是每次折半。
解题代码：
1 // BEGIN CUT HERE 2 /* 3 4 */ 5 // END CUT HERE 6 #line 7 "FoldingPaper2.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 45 int find(int now,int val) 46 { 47 if(now == val) 48 return 0 ; 49 if(val > now/2) 50 return 1; 51 if(now % 2 == 0 ) 52 return 1 + find(now/2,val); 53 return 1 + find(now/2+1,val); 54 } 55 class FoldingPaper2 56 { 57 public: 58 int solve(int W,int H, int A) 59 { 60 int mi = 1e9; 61 int ok = 0 ; 62 for(int i = 1; i <= sqrt(A) + 1;i ++) 63 { 64 if(A % i == 0 ) 65 { 66 if(i <= W && A/i <= H) 67 { 68 mi = min(find(W,i)+find(H,A/i),mi); 69 ok = 1; 70 } 71 if(i <= H && A/i <= W) 72 { 73 mi = min(find(H,i)+find(W,A/i),mi); 74 ok = 1; 75 } 76 } 77 } 78 if(ok) 79 return mi ; 80 else return -1; 81 82 } 83 84 // BEGIN CUT HERE 85 public: 86 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(); } 87 private: 88 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(); } 89 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 << " Expected: "" << Expected << '"' << endl; cerr << " Received: "" << Received << '"' << endl; } } 90 void test_case_0() { int Arg0 = 5; int Arg1 = 3; int Arg2 = 12; int Arg3 = 1; verify_case(0, Arg3, solve(Arg0, Arg1, Arg2)); } 91 void test_case_1() { int Arg0 = 2; int Arg1 = 2; int Arg2 = 3; int Arg3 = -1; verify_case(1, Arg3, solve(Arg0, Arg1, Arg2)); } 92 void test_case_2() { int Arg0 = 4; int Arg1 = 4; int Arg2 = 1; int Arg3 = 4; verify_case(2, Arg3, solve(Arg0, Arg1, Arg2)); } 93 void test_case_3() { int Arg0 = 127; int Arg1 = 129; int Arg2 = 72; int Arg3 = 8; verify_case(3, Arg3, solve(Arg0, Arg1, Arg2)); } 94 void test_case_4() { int Arg0 = 1; int Arg1 = 100000; int Arg2 = 100000; int Arg3 = 0; verify_case(4, Arg3, solve(Arg0, Arg1, Arg2)); } 95 void test_case_5() { int Arg0 = 1; int Arg1 = 1; int Arg2 = 2; int Arg3 = -1; verify_case(5, Arg3, solve(Arg0, Arg1, Arg2)); } 96 97 // END CUT HERE 98 99 }; 100 101 // BEGIN CUT HERE 102 int main() 103 { 104 FoldingPaper2 ___test; 105 ___test.run_test(-1); 106 return 0; 107 } 108 // END CUT HERE