AtCoder：Chocolate Bar（数学）

C - Chocolate Bar

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Time limit : 2sec / Memory limit : 256MB

Score : 400 points

Problem Statement

There is a bar of chocolate with a height of H blocks and a width of W blocks. Snuke is dividing this bar into exactly three pieces. He can only cut the bar along borders of blocks, and the shape of each piece must be a rectangle.

Snuke is trying to divide the bar as evenly as possible. More specifically, he is trying to minimize Smax - Smin, where Smax is the area (the number of blocks contained) of the largest piece, and Smin is the area of the smallest piece. Find the minimum possible value of Smax−Smin.

Constraints

• 2≤H,W≤105

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Input

Input is given from Standard Input in the following format:

H W

Output

Print the minimum possible value of Smax−Smin.

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Sample Input 1

Copy

3 5

Sample Output 1

Copy

0

In the division below, Smax−Smin=5−5=0.

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Sample Input 2

Copy

4 5

Sample Output 2

Copy

2

In the division below, Smax−Smin=8−6=2.

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Sample Input 3

Copy

5 5

Sample Output 3

Copy

4

In the division below, Smax−Smin=10−6=4.

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Sample Input 4

Copy

100000 2

Sample Output 4

Copy

1

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Sample Input 5

Copy

100000 100000

Sample Output 5

Copy

50000

题意：将一个矩形分成三份，使得三份面积的极差最小，输出极差。

思路：割法无非就两种，平行割三刀，或者先割一刀分成两半1和2，再在其中一半横着割一刀分成1，3，4，那么比较这俩情况就行了，对于第二种情况，最小的矩形要么是1要么是3或4，再分这两种情况计算就行了。

# include <bits/stdc++.h>using namespace std; typedef long long LL;int cal(int a, int b){    int c = a/2;    if((LL)a+c > (LL)b*c)         return (int)((LL)a+c-(LL)b*c);    int x = (int)ceil(b*c*1.0/(a+c));    return max((a+c)*x-b*c, (b-x)*(a-2*c));}int cal2(int a, int b){    int c = (int)ceil(a/2.0);    if(a+c > (LL)b*c) return 1e9;    int x = (LL)b*c/(c+a);    return max((int)((LL)b*c-(LL)x*(a+c)), (int)((LL)(b-x)*(2*c-a)));}int cal3(int a, int b){    double x = floor(b/3.0);    double y = ceil(b/3.0);    return (int)(y*a-x*a);}int main(){    int a, b;    scanf("%d%d",&a,&b);    int ans = min(cal(a, b), min(cal(b, a), min(cal2(a, b), min(cal2(b, a), min(cal3(a, b), cal3(b, a))))));    printf("%d\n",ans);    return 0;}