题目大意:n*m的矩形，切成面积相等的n块，要求N 块的长边与短边的比值的最大值最小。n<=10.
初看这道题没有思路，虽然n<=10但是又没说只能切成整数，所以就不知道怎么捉。。二分答案又没想到怎么验证。
事实上，我没有注意到面积相等这个条件。由于只能切（不能拼），所以每次切完的面积必然是n的整数倍，这样就可以搜了。
※注意
double r; int n=5; r=1/n; printf("lf",r);//此时输出0.000000 r=(double)1/(double)n; printf("%lf",r);//此时输出0.200000
View Code
1 /************************************************************** 2 Problem: 1024 3 Language: C++ 4 Result: Accepted 5 Time:676 ms 6 Memory:804 kb 7 ****************************************************************/ 8 9 #include <cstdio> 10 #include <cstdlib> 11 #include <cmath> 12 #include <algorithm> 13 #define bigger(a,b) ((a)>(b)?(a):(b)) 14 #define minner(a,b) ((a)<(b)?(a):(b)) 15 #define INF 999999999; 16 double x,y; 17 int n; 18 double ans; 19 double dfs(double x,double y,int n) 20 { 21 int i; 22 double r,ans=INF; 23 double rtn=bigger(x,y)/minner(x,y); 24 //printf("-------%lf %lf %d %lf\n",x,y,n,rtn); 25 if(n==1) return rtn; 26 r=(double)1/(double)n; 27 for (i=1;i<n;i++) 28 { 29 double a,b; 30 a=bigger(dfs(x*i*r,y,i),dfs(x*(n-i)*r,y,n-i)); 31 b=bigger(dfs(x,y*i*r,i),dfs(x,y*(n-i)*r,n-i)); 32 ans=minner(ans,minner(a,b)); 33 } 34 return ans; 35 } 36 int main() 37 { 38 scanf("%lf%lf%d",&x,&y,&n); 39 double ans=dfs(x,y,n); 40 printf("%lf",ans); 41 return 0; 42 }