一些感悟...

B. Door Frames

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Petya has equal wooden bars of length n. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length a and one top side of length b. A solid (i.e. continuous without breaks) piece of bar is needed for each side.

Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar).

Input

The first line contains a single integer n (1 ≤ n ≤ 1 000) — the length of each wooden bar.

The second line contains a single integer a (1 ≤ a ≤ n) — the length of the vertical (left and right) sides of a door frame.

The third line contains a single integer b (1 ≤ b ≤ n) — the length of the upper side of a door frame.

Output

Print the minimal number of wooden bars with length n which are needed to make the frames for two doors.

Examples

input

Copy

8
1
2

output

Copy

1

input

Copy

5
3
4

output

Copy

6

input

Copy

6
4
2

output

Copy

4

input

Copy

20
5
6

output

Copy

2

Note

In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.

In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed.

CF一道水题。。。一开始用贪心一直过不了，先把代码发一下...

#include <iostream>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <functional>
#include <map>
using namespace std;
int main ()
{
int n;
int a,b;
cin >> n >> a >> b;
int asum = 4;
int bsum = 2;
int ans = 0;
int t = n;
if(a > b)
{
while(1)
{
while(t >= a && asum > 0)
{
t -= a;
asum--;
if(t <= a)
{
// ans++;
break;
}
}
if(t < b && bsum == 0)
{
ans++;
}
while(t >= b && bsum > 0)
{


t -= b;
bsum--;
if(t < b)
{
ans++;
t = n;
break;
}
}
t = n;
if(asum == 0 && bsum == 0)
{
break;
}


}
}
else
{
while(1)
{
while(t >= b && bsum > 0)
{
t -= b;
bsum--;
if(t < b || bsum == 0)
{
break;
}
}
if(t < a && asum == 0)
{
ans++;
}
while(t >= a && asum > 0)
{


t -= a;
asum--;
if(t < a || asum == 0)
{
ans++;
t = n;
break;
}
}
t = n;
if(asum == 0 && bsum == 0)
{
break;
}


}
}
cout << ans << endl;
}


后来看了大神的代码调用了API的随机函数...

#include <cstdio>  
#include <algorithm>  
#include <ctime>  
using namespace std;  


int d[10];  
int n,a,b,i;  
int len,cnt,ans;  


inline int Min(int x, int y){ return x<y?x:y; }  
int main(){  
while (scanf("%d",&n)==1){  
scanf("%d",&a);  
scanf("%d",&b);  
d[1] = d[2] = d[3] = d[4] = a;  
d[5] = d[6] = b;  
ans = 6; cnt = 10;  
for (srand(time(0)), i = 1000; i--; ){  //因为数据量比较小 循环1000次就OK
random_shuffle(d+1,d+7);  //下标从1开始
cnt = len = 0;  
for (int k=1; k<=6; k++) {  
if (len < d[k]) {len = n; cnt++; }  
len -= d[k];  
}  
ans = Min(ans,cnt);
}  
printf("%d\n",ans);  
}  
return 0;  
}  

有些贪心并不是我们想象中的那么简单...