【jzoj3598】【CQOI2014】【数三角形 】

题目大意

解题思路

考虑正难则反，求出所有点组合情况减去三点共线情况，考虑枚举一个矩形，两个端点在对角，第三个在对角线，选取情况是长宽gcd-1，统计答案即可。

code

#include<cmath>#include<cstdio>#include<cstring>#include<algorithm>#define LF double#define LL long long#define Min(a,b) ((a<b)?a:b)#define Max(a,b) ((a>b)?a:b)#define Fo(i,j,k) for(int i=j;i<=k;i++)#define Fd(i,j,k) for(int i=j;i>=k;i--)using namespace std;int const Mxn=1e5;int N,M,Z;int Gcd(int X,int Y){    while(Z=X%Y){        X=Y;        Y=Z;    }    return Y;}int main(){    freopen("d.in","r",stdin);    freopen("d.out","w",stdout);    scanf("%d%d",&N,&M);    if((!N)||(!M)){        printf("0");        return 0;    }    LL NM=1ll*(N+1)*(M+1),Ans=NM*(NM-1)*(NM-2)/6        -1ll*(N+1)*(M+1)*M*(M-1)/6        -1ll*(M+1)*(N+1)*N*(N-1)/6;    Fo(i,2,N)Fo(j,2,M)Ans-=1ll*2*(N-i+1)*(M-j+1)*(Gcd(i,j)-1);    printf("%lld",Ans);    return 0;}