【解题思路】
爆搜，状态f(r,x,y)表示剩下r刀，边长为x和y，对于每个状态枚举切成两块后的长度比或宽度比。复杂度o((n/2)n)。
【参考代码】
1 #include <algorithm> 2 #include <cstdio> 3 #define REP(I,start,end) for(int I=(start);I<=(end);I++) 4 #define PER(I,start,end) for(int I=(start);I>=(end);I--) 5 #define REPs(I,start,end,step) for(int I=(start);I<=(end);I+=(step)) 6 #define PERs(I,start,end,step) for(int I=(start);I>=(end);I-=(step)) 7 #define maxint 32767 8 #define maxlongint 2147483647 9 #define maxint64 9223372036854775807ll 10 using namespace std; 11 typedef unsigned short US; 12 typedef unsigned long UL; 13 typedef long long LL; 14 typedef unsigned long long ULL; 15 inline int getint() 16 { 17 char ch=getchar(); 18 while((ch<'0'||ch>'9')&&ch!='-') 19 ch=getchar(); 20 int result=0; 21 bool impositive=ch=='-'; 22 if(impositive) 23 ch=getchar(); 24 while(ch>='0'&&ch<='9') 25 { 26 result=(result<<3)+(result<<1)+ch-'0'; 27 ch=getchar(); 28 } 29 return impositive?-result:result; 30 } 31 inline int putint(int n) 32 { 33 int result=1; 34 char* sav=new char[20]; 35 bool impositive=n<0; 36 if(impositive) 37 { 38 putchar('-'); 39 n=-n; 40 } 41 sav[0]=n%10+'0'; 42 while(n/=10) 43 sav[result++]=n%10+'0'; 44 PER(i,result-1,0) 45 putchar(sav[i]); 46 delete []sav; 47 return result+impositive; 48 } 49 inline LL getLL() 50 { 51 char ch=getchar(); 52 while((ch<'0'||ch>'9')&&ch!='-') 53 ch=getchar(); 54 LL result=0ll; 55 bool impositive=ch=='-'; 56 if(impositive) 57 ch=getchar(); 58 while(ch>='0'&&ch<='9') 59 { 60 result=(result<<3)+(result<<1)+ch-'0'; 61 ch=getchar(); 62 } 63 return impositive?-result:result; 64 } 65 inline int putLL(LL n) 66 { 67 int result=1; 68 char* sav=new char[20]; 69 bool impositive=n<0; 70 if(impositive) 71 { 72 putchar('-'); 73 n=-n; 74 } 75 sav[0]=n%10+'0'; 76 while(n/=10) 77 sav[result++]=n%10+'0'; 78 PER(i,result-1,0) 79 putchar(sav[i]); 80 delete []sav; 81 return result+impositive; 82 } 83 template<typename integer> inline int read_int(integer &n) 84 { 85 char ch=getchar(); 86 while((ch<'0'||ch>'9')&&ch!='-') 87 ch=getchar(); 88 int result=n=integer(0); 89 bool impositive=ch=='-'; 90 if(impositive) 91 ch=getchar(); 92 while(ch>='0'&&ch<='9') 93 { 94 n=(n<<3)+(n<<1)+integer(ch-'0'); 95 result++; 96 ch=getchar(); 97 } 98 if(impositive) 99 { 100 n=-n; 101 result++; 102 } 103 return result; 104 } 105 template<typename integer> inline int write_int(integer n) 106 { 107 int result=1; 108 char* sav=new char[20]; 109 bool impositive=n<0; 110 if(impositive) 111 { 112 putchar('-'); 113 n=-n; 114 } 115 sav[0]=n%10+'0'; 116 while(n/=10) 117 sav[result++]=n%10+'0'; 118 PER(i,result-1,0) 119 putchar(sav[i]); 120 delete []sav; 121 return result+impositive; 122 } 123 template<typename T> inline bool getmin(T &target,T pattern) 124 { 125 bool can=pattern<target; 126 if(can) 127 target=pattern; 128 return can; 129 } 130 template<typename T> inline bool getmax(T &target,T pattern) 131 { 132 bool can=pattern>target; 133 if(can) 134 target=pattern; 135 return can; 136 } 137 //==========================================Header Template================================================ 138 #include <iomanip> 139 #include <iostream> 140 double DFS(int rest,double _x,double _y) 141 { 142 if(rest==1) 143 return _x>_y?_x/_y:_y/_x; 144 double result=maxlongint; 145 REP(i,1,rest>>1) 146 { 147 long double x0=_x/rest,y0=_y/rest,_r=rest-i; 148 getmin(result,min(max(DFS(i,x0*i,_y),DFS(_r,x0*_r,_y)),max(DFS(i,_x,y0*i),DFS(_r,_x,y0*_r)))); 149 } 150 return result; 151 } 152 int main() 153 { 154 int x=getint(),y=getint(),n=getint(); 155 cout<<setiosflags(ios::fixed)<<setprecision(6)<<DFS(n,x,y)<<endl; 156 return 0; 157 }