像我這種弱比，十一不能回家，中秋不能回家的弱比，才會在這里默默的刷題，我擦
昨天zoj月賽，坑死爹了
ZOJ Problem Set - 3647
Gao the Grid
Time Limit: 2 Seconds Memory Limit: 65536 KB
A n * m grid as follow:
Count the number of triangles, three of whose vertice must be grid-points.
Note that the three vertice of the triangle must not be in a line(the right picture is not a triangle).
Input
The input consists of several cases. Each case consists of two positive integers n and m (1 ≤ n, m ≤ 1000).
Output
For each case, output the total number of triangle.
Sample Input
1 1
2 2
Sample Output
4
76
hint
hint for 2nd case: C(9, 3) - 8 = 76
yy出了想法，想法有些問題，還有搞的重復了好多
然后今天看了woshi250hua大牛的解題報告，修改了代碼
然后自認為沒錯了，還是一直wa，
然后我就改了改輸出 %I64d ->%lld
就過了，坑爹啊
中秋節木有妹子陪，也木有月餅吃，哎。。。苦逼的日子
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <cassert>
#include <iostream>
#include <sstream>
#include <fstream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <algorithm>
#include <iomanip>
#define maxn 1010
#define LL long long
using namespace std;
LL f[maxn][maxn];
LL num[maxn];
LL totpoint,sum;
LL n,m;
int gcd(int a,int b)
{
if(b==0) return a;
else return gcd(b,a%b);
}
void work()
{
memset(num,0,sizeof(num));
num[0]=num[1]=num[2]=0;
for(int i=3; i<=1002; i++)
if(i-1&&i-2)
num[i] = i * (i - 1) * (i - 2) / 6;
}
int main()
{
memset(f,0,sizeof(f));
for(int i=1; i<=1002; i++)
{
for(int j=1; j<=1002; j++)
f[i][j]=gcd(i,j)-1;
}
work();
while(scanf("%I64d%I64d",&n,&m)!=EOF)
{
totpoint= (n+1) * (m+1);
sum = totpoint * ( totpoint - 1 ) * (totpoint - 2) / 6 ;
sum -= (m+1)*(num[n+1]);
sum -= (n+1)*(num[m+1]);
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
{
sum -= f[i][j]*(n+1-i)*(m+1-j)*2;
}
printf("%lld\n",sum);
}
return 0;
}