codeforces #313(div 2)

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B. Gerald is into Art

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Gerald bought two very rare paintings at the Sotheby's auction and he now wants to hang them on the wall. For that he bought a special board to attach it to the wall and place the paintings on the board. The board has shape of ana₁ × b₁
rectangle, the paintings have shape of aa₂ × b₂ anda₃ × b₃
rectangles.

Since the paintings are painted in the style of abstract art, it does not matter exactly how they will be rotated, but still, one side of both the board, and each of the paintings must be parallel to the floor. The paintings can touch each other and the
edges of the board, but can not overlap or go beyond the edge of the board. Gerald asks whether it is possible to place the paintings on the board, or is the board he bought not large enough?

Input

The first line contains two space-separated numbers a₁ andb₁ — the sides of the board. Next two lines contain numbersa₂, b₂, a₃
andb₃ — the sides of the paintings. All numbersa_i, b_i in the
input are integers and fit into the range from1 to
1000.

Output

If the paintings can be placed on the wall, print "YES" (without the quotes), and if they cannot, print "NO" (without the quotes).

Sample test(s)

Input

3 2
1 3
2 1

Output

YES

Input

5 5
3 3
3 3

Output

Input

4 2
2 3
1 2

Output

YES

Note

That's how we can place the pictures in the first test:

And that's how we can do it in the third one.

题目要求把两幅画放到一块黑板上，要求两幅画不重叠不越界。所以能够先将一幅画放上去。在剩下的区域里面能够有两个位置去放，暴力枚举各种情况就可以。当中要注意第一幅画本身是否能放上去。

#include<stdio.h>
#include<string.h>
int main()
{
int a1,b1,a2,b2,a3,b3,i,j,k;
while(scanf("%d%d",&a1,&b1)!=EOF)
{
scanf("%d%d%d%d",&a2,&b2,&a3,&b3);
int x1=a1-a2;
int x2=a1-b2;
int y1=b1-b2;
int y2=b1-a2;
int f=0; if((x1>=a3&&b1>=b3&&b1>=b2)||(x1>=b3&&b1>=a3&&b1>=b2)){
printf("YES\n");continue;
}
if((y1>=b3&&a1>=a3&&a1>=a2)||(y1>=a3&&a1>=b3&&a1>=a2)){
printf("YES\n");continue;
}
if((x2>=a3&&b1>=b3&&b1>=a2)||(x2>=b3&&b1>=a3&&b1>=a2)){
printf("YES\n");continue;
}
if((y2>=a3&&a1>=b3&&a1>=b2)||(y2>=b3&&a1>=a3&a1>=b2)){
printf("YES\n");continue;
}
else printf("NO\n");
}
}

C. Gerald's Hexagon

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Gerald got a very curious hexagon for his birthday. The boy found out that all the angles of the hexagon are equal to.
Then he measured the length of its sides, and found that each of them is equal to an integer number of centimeters. There the properties of the hexagon ended and Gerald decided to draw on it.

He painted a few lines, parallel to the sides of the hexagon. The lines split the hexagon into regular triangles with sides of 1 centimeter. Now Gerald wonders how many triangles he has got. But there were so many of them that Gerald lost the track of his
counting. Help the boy count the triangles.

Input

The first and the single line of the input contains 6 space-separated integersa₁, a₂, a₃, a₄, a₅
anda₆ (1 ≤ a_i ≤ 1000) — the lengths of the sides of the hexagons in centimeters in the clockwise order. It
is guaranteed that the hexagon with the indicated properties and the exactly such sides exists.

Output

Print a single integer — the number of triangles with the sides of one 1 centimeter, into which the hexagon is split.

Sample test(s)

Input

1 1 1 1 1 1

Output

Input

1 2 1 2 1 2

Output

Note

This is what Gerald's hexagon looks like in the first sample:

And that's what it looks like in the second sample:

这个是数学题，推一下都能够推出来。题目大意,给定六边形的边长，要求算出当中有多少个等腰三角形。图形能够确定是由等腰三角形组成的。

假设画一绘图就能发现，假设将图形补成一个大的等腰三角形，那么小三角形的数目就是左側（或右側）三条边加起来以后的平方。由于对于一个边长为n 的等腰三角形来说。里面的小三角形是有n^2个的。所以题目中要求的三角形数量就是大的减去边上三个小的。

#include<stdio.h>
int s(int i)
{
return i*i;
}
int main()
{
int a[6],ans;
while(~scanf("%d",&a[0]))
{
for(int i=1;i<6;i++)
scanf("%d",&a[i]);
ans=a[0]+a[1]+a[2];
printf("%d\n",s(ans)-s(a[0])-s(a[2])-s(a[4]));
}
}

D. Equivalent Strings

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings
a and b of equal length are called
equivalent in one of the two cases:

They are equal.
If we split string a into two halves of the same size
a₁ and
a₂, and string
b into two halves of the same size b₁ and
b₂, then one of the following is correct:
1. a₁ is equivalent to
  b₁, and
  a₂ is equivalent to
  b₂
2. a₁ is equivalent to
  b₂, and
  a₂ is equivalent to
  b₁

As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.

Gerald has already completed this home task. Now it's your turn!

Input

The first two lines of the input contain two strings given by the teacher. Each of them has the length from
1 to 200 000 and consists of lowercase English letters. The strings have the same length.

Output

Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.

Sample test(s)

Input

aaba
abaa

Output

YES

Input

aabb
abab

Output

Note

In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa".
"aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab"
= "a" + "b", "ba" = "b" + "a".

In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb"
is equivalent only to itself and to string "bbaa".

题目大意：推断两个字符串是否“同样”。“同样”仅仅要满足下列条件之中的随意一个：

They are equal.
If we split string a into two halves of the same size
a₁ and
a₂, and string
b into two halves of the same size b₁ and
b₂, then one of the following is correct:
1. a₁ is equivalent to
  b₁, and
  a₂ is equivalent to
  b₂
2. a₁ is equivalent to
  b₂, and
  a₂ is equivalent to
  b₁

因为我拆分字符串的时候一定要等分，所以假设字符串是奇数的，直接strcmp比較就可以。

假设是偶数，那么久要拆分了。这里注意，我能够一直拆分直到被拆分的字符串变成奇数。

拆分过程中，假设有一组满足条件。那就满足条件了。所以能够调用递归函数。

#include<stdio.h>
#include<string.h>
char a[200005],b[200005];
bool panduan(char * s1,char *s2,int l)
{if(strncmp(s1,s2,l)==0)return true; //这里用了strncmp函数。即比較s1,s2前l个字符是否同样。
if(l%2==1)return false;
int i,j;
/*char s3[100005],s4[100005];
char s5[100005],s6[100005];
for(i=0;i<l/2;i++)
{
s3[i]=s1[i];
s5[i]=s2[i];
}
for(i=l/2;i<l;i++)
{
s4[i-l/2]=s1[i];
s6[i-l/2]=s2[i];
}
*/ //这里的部分是我第一次写的时候T的代码，没实用strncmp函数。应该是由于递归调用过多，for循环用的也特别多。if(panduan(s1,s2+l/2,l/2)&&panduan(s1+l/2,s2,l/2))return true;
if(panduan(s1,s2,l/2)&&panduan(s1+l/2,s2+l/2,l/2))return true; //注意这一行与上一行这题里面换不了，一换就T了。预计是数据问题。//假设实在要换，能够手写strncmp函数。可能能够过，我没试过。return false;}
int main()
{
int i,j,k,l;
while(scanf("%s%s",a,b)!=EOF)
{
l=strlen(a);
if(l%2==1){
if(strcmp(a,b)==0)printf("YES\n");
else printf("NO\n");
}
else {
if(panduan(a,b,strlen(a)))printf("YES\n");
else printf("NO\n");
}
}
}