BZOJ1005 HNOI2008明明的煩惱(prufer+高精度)
阿新 • • 發佈:2019-02-17
const clu sum lld ans else con pre ||
每個點的度數=prufer序列中的出現次數+1,所以即每次選一些位置放上某個點,答案即一堆組合數相乘。記一下每個因子的貢獻分解一下質因數高精度乘起來即可。
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll long long #define N 1010 char getc(){char c=getchar();while ((c<‘A‘||c>‘Z‘)&&(c<‘a‘||c>‘z‘)&&(c<‘0‘||c>‘9‘)) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;char c=getchar(); while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();} while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } int n,a[N],f[N],g[N],ans[N<<2]; int main() { #ifndef ONLINE_JUDGE freopen("bzoj1005.in","r",stdin); freopen("bzoj1005.out","w",stdout); const char LL[]="%I64d\n"; #else const char LL[]="%lld\n"; #endif n=read(); for (int i=1;i<=n;i++) a[i]=read()-1; for (int i=1;i<=n;i++) if (a[i]==-1) {cout<<0;return 0;} int sum=0;for (int i=1;i<=n;i++) if (a[i]==0) sum+=a[i]; if (sum>n-2){cout<<0;return 0;}sum=n-2;int cnt=0; for (int i=1;i<=n;i++) if (a[i]>=0) { for (int j=sum-a[i]+1;j<=sum;j++) f[j]++; for (int j=1;j<=a[i];j++) f[j]--; sum-=a[i]; } else cnt++; f[cnt]+=sum; for (int i=2;i<=n;i++) { int x=i; for (int j=2;j<=x;j++) while (x%j==0) g[j]+=f[i],x/=j; if (x>1) g[x]+=f[i]; } ans[1]=1;int len=1; for (int i=2;i<=n;i++) while (g[i]--) { for (int j=1;j<=len;j++) ans[j]*=i; for (int j=1;j<=len;j++) ans[j+1]+=ans[j]/10,ans[j]%=10; while (ans[len+1]) len++,ans[len+1]+=ans[len]/10,ans[len]%=10; } for (int i=len;i>=1;i--) printf("%d",ans[i]); return 0; }
BZOJ1005 HNOI2008明明的煩惱(prufer+高精度)