题解
莫比乌斯反演（可我式子都没列出来。。。直接写了个n^4+剪枝没想到有65分）
我推出来的式子不太一样，但时间复杂度使一样的。。。
代码：
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 100005
const int mod=323232323;
int gcd(int x,int y){return !y?x:gcd(y,x%y);}
int fac[N],inv[N],prime[N],mu[N],tot;
bool vis[N];
void shai()
{
int n=100000,i,j;
vis[1]=1;mu[1]=1;
for(i=2;i<=n;i++){
if(!vis[i]){
prime[++tot]=i;
mu[i]=mod-1;
}
for(j=1;j<=tot;j++){
int tmp=i*prime[j];
if(tmp>n)break;
vis[tmp]=1;
if(i%prime[j]==0){mu[tmp]=0;break;}
mu[tmp]=mod-mu[i];
}
}
fac[0]=fac[1]=inv[0]=inv[1]=1;
for(i=2;i<=n;i++){
fac[i]=1ll*fac[i-1]*i%mod;
inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
}
for(i=2;i<=n;i++)
inv[i]=1ll*inv[i-1]*inv[i]%mod;
}
int C(int x,int y)
{
if(y<0||x<y)return 0;
return 1ll*fac[x]*inv[y]%mod*inv[x-y]%mod;
}
int F1(int x)
{
return 1ll*x*(x+1)/2%mod;
}
int main()
{
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
int n,m,K,d,g;
int ans=0;shai();
scanf("%d%d%d",&n,&m,&K);
if(K==1){printf("%d",int(1ll*n*m%mod));return 0;}
for(g=1;g<=min(n,m);g++){
int sum=0,lim=min(n/g,m/g);
for(d=1;d<=lim;d++)
sum=(sum+1ll*mu[d]*(1ll*(n/(d*g))*n%mod+1ll*mod-1ll*d*g*F1(n/(d*g))%mod)%mod
*(1ll*(m/(d*g))*m%mod+1ll*mod-1ll*d*g*F1(m/(d*g))%mod)%mod)%mod;
ans=(1ll*ans+1ll*C(g-1,K-2)*sum)%mod;
}
ans=2ll*ans%mod;
ans=(ans+1ll*C(n,K)*m%mod+1ll*C(m,K)*n%mod)%mod;
printf("%d",ans);
}
题解
代码：还没写
题解
代码：
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
inline int gi()
{
char c;int num=0,flg=1;
while((c=getchar())<'0'||c>'9')if(c=='-')flg=-1;
while(c>='0'&&c<='9'){num=num*10+c-48;c=getchar();}
return num*flg;
}
#define N 10005
const int mod=323232323;
int n,a[N],fa[N],siz[N],sum[N],f[N][N];
int main()
{
freopen("c.in","r",stdin);
freopen("c.out","w",stdout);
int i,j,k,x,y;
n=gi();
for(i=1;i<=n;i++){
a[i]=gi()+i-1;
if(a[i]>n){printf("0\n");return 0;}
}
a[1]=n;
for(i=n;i>=1;i--){
fa[i]=i;siz[i]=1;
a[i]=a[fa[a[i]]];
f[i][siz[i]]=1;
for(j=i+1;j<=a[i];j=a[j]+1){
sum[siz[i]+1]=0;
for(k=siz[i];k>=1;k--){
sum[k]=(sum[k+1]+f[i][k])%mod;
f[i][k]=0;
}
for(x=1;x<=siz[i];x++)
for(y=1;y<=siz[j];y++)
f[i][x+y]=(1ll*f[i][x+y]+1ll*sum[x]*f[j][y])%mod;
siz[i]+=siz[j];
}
for(j=i+1;j<=a[i];j++) fa[j]=i;
}
int ans=0;
for(i=1;i<=n;i++)ans=(ans+f[1][i])%mod;
printf("%d\n",ans);
}
