Codeforces Round #313 (Div. 2) 560C Gerald's Hexagon(脑洞)

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 integers a1, a2, a3, a4, a5 and a6 (1 ≤ ai ≤ 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

6

input

1 2 1 2 1 2

output

13

Note

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



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



题目链接:点击打开链接

给出六边形的六条边, 问你六边形由多少个小正三角形组成.

将六边形补成一个正大三角形, 能够发现大三角形边长为(a[1] + a[2] + a[3]) * (a[1] * a[2] * a[3]), 即一个边长为n的正三角形包括n * n个

小的正三角形, 再减去补的三个小正三角形包括的个数就可以.

AC代码:

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "queue"
#include "stack"
#include "cmath"
#include "utility"
#include "map"
#include "set"
#include "vector"
#include "list"
#include "string"
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 10;
ll a[MAXN];
int main(int argc, char const *argv[])
{
for(int i = 1; i <= 6; ++i)
scanf("%lld", &a[i]);
printf("%lld\n", (a[1] + a[2] + a[3]) * (a[1] + a[2] + a[3]) - a[1] * a[1] - a[3] * a[3] - a[5] * a[5]);
return 0;
}