标签:lin geo href ++ for bzoj open std cst
http://www.lydsy.com/JudgeOnline/problem.php?id=3505 (题目链接)
给定一个n*m的网格,请计算三点都在格点上的三角形共有多少个。
$${ans=平面中选三个点的方案数-三点共线的方案数}$$
$${ans=C_{(n+1)*(m+1)}^{3}-(n+1)*C_{m+1}^{3}-(m+1)*C_{n+1}^{3}-斜的三点共线的方案数}$$
斜的三点共线方案数不会求。。左转题解:http://blog.csdn.net/zhb1997/article/details/38474795
LL
// bzoj3505 #include<algorithm> #include<iostream> #include<cstdlib> #include<cstring> #include<cstdio> #include<cmath> #include<queue> #define LL long long #define inf 10000000 #define Pi acos(-1.0) #define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout); using namespace std; int n,m; LL c[2000010][4]; int gcd(int a,int b) { return b==0 ? a : gcd(b,a%b); } int main() { scanf("%d%d",&n,&m); for (int i=0;i<=(n+1)*(m+1);i++) c[i][0]=1; for (int i=1;i<=(n+1)*(m+1);i++) for (int j=1;j<=min(3,i);j++) c[i][j]=c[i-1][j-1]+c[i-1][j]; LL ans=c[(n+1)*(m+1)][3]-(n+1)*c[m+1][3]-(m+1)*c[n+1][3]; for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) { LL x=gcd(i,j)+1; if (x>2) ans-=(x-2)*2*(n-i+1)*(m-j+1); } printf("%lld",ans); return 0; }
标签:lin geo href ++ for bzoj open std cst
原文地址:http://www.cnblogs.com/MashiroSky/p/6217010.html