BZOJ P1005:[HNOI2008]明明的烦恼

脑残题目打了半天打不出来

这高精打的我要吐血(其实是ctrlC+V的)

我决定不A这题了

但是思路还是讲讲

这题还是肥肠显然的prufer数列的应用嘛

我的prufer数列一共有n-2个元素嘛(n为节点个数)

那么我已经确定的有sigma(dushu[i]-1)个嘛

那么一共的方案数就有C(n-2,sigma(dushu[i]-1))嘛

但是还有重复的情况嘛

那么就除掉tot!/(d(s)-1)嘛

然后还有剩下的一些空嘛

这些空可以被所有度数为-1的节点填上嘛

那么就是快速幂k^(n-2-tot)嘛

下面是我WA的代码

有兴趣的可以帮我找找错误

#include<iostream>#include<fstream>#include<algorithm>#include<cmath>#include<cstring>using namespace std;const int MAXN = 4100;int n;int d[10003];int k;int p[2][1003];void cal(int x,int k){  for(int i=2;;i++){        while(x%i==0){p[k][i]++,x/=i;}        if(x==1){        break;}    }}struct bign{      int len,s[MAXN];      bign (){          memset(s,0,sizeof(s));          len=1;      }      bign (int num) { *this = num; }      bign (const char *num) { *this = num; }      bign operator = (const int num){          char s[MAXN];          sprintf(s, "%d", num);          *this = s;          return *this;      }      bign operator = (const char *num){          for(int i = 0; num[i] == '0'; num++) ;  //去前导0          len = strlen(num);          for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';          return *this;      }      bign operator + (const bign &b) const{          bign c;  //+        c.len = 0;          for(int i = 0, g = 0; g || i < max(len, b.len); i++){              int x = g;              if(i < len) x += s[i];              if(i < b.len) x += b.s[i];              c.s[c.len++] = x % 10;              g = x / 10;          }          return c;      }      bign operator += (const bign &b){          *this = *this + b;          return *this;      }      void clean(){          while(len > 1 && !s[len-1]) len--;      }      bign operator * (const bign &b) {          bign c;  //*        c.len = len + b.len;          for(int i = 0; i < len; i++){              for(int j = 0; j < b.len; j++){                  c.s[i+j] += s[i] * b.s[j];              }          }          for(int i = 0; i < c.len; i++){              c.s[i+1] += c.s[i]/10;              c.s[i] %= 10;          }          c.clean();          return c;      }      bign operator *= (const bign &b){          *this = *this * b;          return *this;      }      bign operator - (const bign &b){          bign c;          c.len = 0;          for(int i = 0, g = 0; i < len; i++){              int x = s[i] - g;              if(i < b.len) x -= b.s[i];              if(x >= 0) g = 0;              else{                  g = 1;                  x += 10;              }              c.s[c.len++] = x;          }          c.clean();          return c;      }  bign operator -= (const bign &b){          *this = *this - b;          return *this;      }      bign operator /(const bign &b){          bign c,f=0;          for(int i=len-1;i >= 0; i--){              f = f*10;              f.s[0] = s[i];              while(f >= b){                  f -= b;                  c.s[i]++;              }          }          c.len = len;          c.clean();          return c;      }      bign operator /=(const bign &b){          *this=*this/b;          return *this;      }      bign operator %(const bign &b){          bign r=*this/b;          r=*this-r*b;          return r;      }      bign operator %=(const bign &b){          *this=*this%b;          return *this;      }      bool operator <(const bign &b){          if(len!=b.len) return len<b.len;          for(int i=len-1;i>=0;i--){              if(s[i]!=b.s[i]) return s[i]<b.s[i];          }          return false;      }      bool operator > (const bign &b){          if(len!=b.len) return len>b.len;          for(int i=len-1;i>=0;i--){              if(s[i]!=b.s[i]) return s[i]>b.s[i];          }          return false;      }      bool operator ==(const bign &b){        return !(*this>b)&&!(*this<b);      }      bool operator != (const bign &b){          return !(*this == b);      }      bool operator <= (const bign &b){          return *this < b || *this == b;      }      bool operator >= (const bign &b){          return *this > b || *this == b;      }      string str() const{          string res="";          for(int i=0; i < len; i++) res = char(s[i]+'0') + res;          return res;      }  }ans;istream& operator >>(istream &in, bign &x){      string s;      in>>s;      x=s.c_str();      return in;  }ostream& operator <<(ostream &out, const bign &x){      out<<x.str();      return out;  }bign qpow(bign x,int y){if(y==1){return x;}else{bign fanhui=qpow(x,y/2);fanhui=fanhui*fanhui;if(y%2==1){fanhui*=x;}return fanhui;}}int main(){cin>>n;int tot=0;ans=1;for(int i=1;i<=n;i++){cin>>d[i];if(d[i]==-1){k++;continue;}tot+=(d[i]-1);}for(int i=2;i<=tot;i++){cal(i,0);cal(i,0);}for(int i=1;i<=n;i++){for(int j=2;j<d[i];j++){cal(i,1);}}bign x=k;for(int i=2;i<=n-2;i++){cal(i,1);}for(int i=2;i<=tot-n+2;i++){cal(i,1);}for(int i=2;i<=1000;i++){if(p[0][i]>p[1][i]){ans*=qpow(i,p[0][i]-p[1][i]);}if(p[1][i]>p[0][i]){ans/=qpow(i,p[1][i]-p[0][i]);}}ans*=qpow(x,n-2-tot);cout<<ans<<endl;return 0;}