链接：http://codeforces.com/problemset/problem/215/C
C. Crosses
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
There is a board with a grid consisting of n rows and m columns,
the rows are numbered from 1 from top to bottom and the columns are numbered from 1 from
left to right. In this grid we will denote the cell that lies on row number i and column number j as (i,?j).
A group of six numbers (a,?b,?c,?d,?x0,?y0),
where 0?≤?a,?b,?c,?d, is a cross, and there
is a set of cells that are assigned to it. Cell (x,?y)belongs to this set if at
least one of two conditions are fulfilled:
- |x0?-?x|?≤?a and |y0?-?y|?≤?b
- |x0?-?x|?≤?c and |y0?-?y|?≤?d
The picture shows the
cross (0,?1,?1,?0,?2,?3) on the grid 3?×?4.
Your task is to find the number of different groups of six numbers, (a,?b,?c,?d,?x0,?y0) that
determine the crosses of an area equal to s, which are placed entirely on the grid. The cross is placed entirely on the grid, if any
of its cells is in the range of the grid (that is for each cell (x,?y) of the cross 1?≤?x?≤?n; 1?≤?y?≤?m holds).
The area of the cross is the number of cells it has.
Note that two crosses are considered distinct if the ordered groups of six numbers that denote them are distinct, even if these crosses coincide as sets of points.
Input
The input consists of a single line containing three integers n, m and s (1?≤?n,?m?≤?500, 1?≤?s?≤?n·m).
The integers are separated by a space.
Output
Print a single integer — the number of distinct groups of six integers that denote crosses with area s and that are fully placed on then?×?m grid.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64dspecifier.
Sample test(s)
input
2 2 1
output
4
input
3 4 5
output
4
Note
In the first sample the sought groups of six numbers are: (0,?0,?0,?0,?1,?1), (0,?0,?0,?0,?1,?2), (0,?0,?0,?0,?2,?1), (0,?0,?0,?0,?2,?2).
In the second sample the sought groups of six numbers are: (0,?1,?1,?0,?2,?2), (0,?1,?1,?0,?2,?3), (1,?0,?0,?1,?2,?2), (1,?0,?0,?1,?2,?3).
题意：
给你n*m矩阵，问有多少个不同的 (a,?b,?c,?d,?x0,?y0)
面积等于s。
- |x0?-?x|?≤?a and |y0?-?y|?≤?b
- |x0?-?x|?≤?c and |y0?-?y|?≤?d
满足这个条件 相当于两个一x0 ，y0 为中心的，边长全为奇数的矩形并。
一个矩形长为2a+1，宽为2b+1 另一个是(2*c+1) * (2*d+1)
做法：
枚举其中一个矩形的长和宽。
如果面积超过s显然不行。
如果等于s，那么另一个肯定比它小或者相等。
ans+=(n-i/2*2)*(m-j/2*2)* (((i/2+1)*(j/2+1)-1)*2+1);
(n-i/2*2)*(m-j/2*2) 这个算有多少位子 可以做为矩形 中心
(((i/2+1)*(j/2+1)-1)*2+1) 枚举那个小的边长
如果小于s的话，枚举另一个矩阵的长度，长度小于i
然后计算宽度。
宽度如果小于m并且也是奇数
ans+=(n-i/2*2)*(m-wid/2*2)*2; 枚举中心点可以在的位置，因为两个矩形不同所以abcd可以对换 所以*2
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <malloc.h>
#include <ctype.h>
#include <math.h>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#include <stack>
#include <queue>
#include <vector>
#include <deque>
#include <set>
#include <map>
int main()
{
int n,m,s;
scanf("%d%d%d",&n,&m,&s);
__int64 ans=0;
for(int i=1;i<=n;i+=2)//枚举比较长的长方形 长度
{
for(int j=1;j<=m;j+=2)
{
if(i*j>s)
continue;
else if(i*j==s)//一个包含另一个
{
ans+=(n-i/2*2)*(m-j/2*2)* (((i/2+1)*(j/2+1)-1)*2+1);
// 位子*枚举边长
}
else
{
for(int len=1;len<i;len+=2)//长小的
{
if((s-i*j)%len==0)
{
int wid=(s-i*j)/(len)+j;
//长的 j
//宽的 wid
if(wid<=m&&(wid&1))
ans+=(n-i/2*2)*(m-wid/2*2)*2;
//确定位置 因为不同 所以abcd可以换 *2
}
}
}
}
}
printf("%I64d\n",ans);
scanf("%d%d%d",&n,&m,&s);
return 0;
}
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