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

weixin_34015860發表於2017-08-05
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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:



題目連結:點選開啟連結

給出六邊形的六條邊, 問你六邊形由多少個小正三角形組成.

將六邊形補成一個正大三角形, 能夠發現大三角形邊長為(a[1] + a[2] + a[3]) * (a[1] * a[2] * a[3]), 即一個邊長為n的正三角形包括n * n個

小的正三角形, 再減去補的三個小正三角形包括的個數就可以.

AC程式碼:

#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;
}


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