nowcoder19934 [CQOI2014]数三角形
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题解
答案等于C(n+1)(m+1)3减去三点共线的情况数
fi,j表示以格点(i,j)为右下角的点时,选三点共线的方案数
fi,j先赋值为fi−1,j+fi,j−1−fi−1,j−1,这样还剩下一种没算,就是以(0,0)为左上角的点,以(i,j)为右下角的点的情况,这种情况的个数为gcd(i,j)−1
综上,fi,j=fi−1,j+fi,j−1−fi−1,j−1+gcd(i,j)−1
根据对称性,答案等于C(n+1)(m+1)3−2×∑fij+(n+1)Cm+13+(m+1)Cn+13
代码
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(_,__) for(_=1;_<=(__);_++)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
ll c, f(1);
for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
for(;isdigit(c);c=getchar())x=x*10+c-0x30;
return f*x;
}
ll calc(ll n){return n*(n-1)*(n-2)/6;}
ll f[maxn][maxn];
int main()
{
ll i, j, n, m, ans;
n=read(), m=read();
ans = calc((n+1)*(m+1));
for(i=0;i<=n;i++)for(j=0;j<=m;j++)
{
if(i)f[i][j] += f[i-1][j];
if(j)f[i][j] += f[i][j-1];
if(i and j)f[i][j] -= f[i-1][j-1];
f[i][j] += max(0ll,__gcd(i,j)-1);
ans -= 2*f[i][j];
}
ans += (n+1)*calc(m+1) + (m+1)*calc(n+1);
printf("%lld",ans);
return 0;
}
*ACoder*
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标签:gcd,ll,nowcoder19934,fi,3C,CQOI2014,三角形,include,getchar 来源: https://blog.csdn.net/FSAHFGSADHSAKNDAS/article/details/104086263