洛谷 P2158 [SDOI2008]仪仗队 && 洛谷 P1447 [NOI2010]能量采集

https://www.luogu.org/problemnew/show/P2158

以人所在位置为(0,0)建立坐标系，

显然除了(0,1)和(1,0)外，可以只在坐标(x,y)的gcd(x,y)==1时统计贡献。。因为如果gcd(x,y)==g而g不等于1，那么会在(x/g,y/g)处统计贡献

1要特判。。

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<vector>
 using namespace std;
 #define fi first
 #define se second
 #define mp make_pair
 #define pb push_back
 typedef long long ll;
 typedef unsigned long long ull;
 typedef pair<int,int> pii;
 #define N 100100
 ll prime[N+],len,mu[N+];
 bool nprime[N+];
 ll n,ans;
 ll F(ll x)    {return (n/x)*(n/x);}
 int main()
 {
     ll i,j;
     mu[]=;
     for(i=;i<=N;i++)
     {
         if(!nprime[i])    prime[++len]=i,mu[i]=-;
         for(j=;j<=len&&i*prime[j]<=N;j++)
         {
             nprime[i*prime[j]]=;
             if(i%prime[j]==)    {mu[i*prime[j]]=;break;}
             else    mu[i*prime[j]]=-mu[i];
         }
     }
     scanf("%lld",&n);
     if(n==)    {puts("");return ;}
     n--;
     for(i=;i<=n;i++)    ans+=mu[i]*F(i);
     printf("%lld",ans+);
     return ;
 }

---

https://www.luogu.org/problemnew/show/P1447

一样的。。。

答案就是$\sum_{i=1}^n\sum_{j=1}^m{(2*gcd(i,j)-1)}$

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<vector>
 using namespace std;
 #define fi first
 #define se second
 #define mp make_pair
 #define pb push_back
 typedef long long ll;
 typedef unsigned long long ull;
 typedef pair<int,int> pii;
 #define N 100000
 ll prime[N+],len,mu[N+];
 bool nprime[N+];
 ll n,m,ans,a2;
 ll F(ll x)    {return (m/x)*(n/x);}
 int main()
 {
     ll i,j,k;
     mu[]=;
     for(i=;i<=N;i++)
     {
         if(!nprime[i])    prime[++len]=i,mu[i]=-;
         for(j=;j<=len&&i*prime[j]<=N;j++)
         {
             nprime[i*prime[j]]=;
             if(i%prime[j]==)    {mu[i*prime[j]]=;break;}
             else    mu[i*prime[j]]=-mu[i];
         }
     }
     scanf("%lld%lld",&n,&m);
     if(n>m)    swap(n,m);
     for(k=;k<=m;k++)
     {
         a2=;
         for(i=;i<=m/k;i++)
             a2+=mu[i]*F(i*k);
         ans+=a2*k;
     }
     printf("%lld",ans*-n*m);
     return ;
 }

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http://210.33.19.103/contest/776/problem/2

完全不会。。