Noi 10 even test DAY3 T1

Source: Internet
Author: User

So I thought for a long time in this two-forced exam, I was so weak ...

First, since ANS is i* (i-1)/2, all that is required is the weight of each number of possible occurrences * of this number.

We find that the essence of each number is the same, we remember a sum of sums, so as long as the statistics are OK.

We abstract each selection into an edge, each state as a point, thus constituting a root tree.

We only consider the 1 contribution to the answer. To calculate the contribution of the current merge to the answer at each level, that is, to know that I have chosen to merge 1 or 1 of the unicom blocks in this node, how many leaf nodes can I cover?

Then it becomes an O (n) combinatorial math problem. The factorial and factorial inverse of the combined number can be preprocessed by O (n).

1#include <cstdio>2#include <iostream>3#include <cmath>4#include <cstring>5#include <algorithm>6 #definell Long Long7 intans=0, a[200005],n,cnt[ -],size[ -];8 Constll mod=1000000007;9ll sum,jcny[200005],jc[200005],f[200005],son[200005],num[200005];Ten intRead () { One     intt=0, f=1;CharCh=GetChar (); A      while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();} -      while('0'<=ch&&ch<='9') {t=t*Ten+ch-'0'; ch=GetChar ();} -     returnt*F; the } - BOOLcmpConst int&a,Const int&b) { -     returnA>b; - } + ll gcd (ll A,ll b) { -     if(b==0)returnA; +     Else returnGCD (b,a%b); A } at voidEXGCD (ll a,ll b,ll &x,ll &y) { -     if(b==0){ -x=1; y=0; -         return; -     } -EXGCD (b,a%b,x,y); inll t=x; -x=y; toY=t-(A/b) *y; + } - voidinit () { thejc[0]=1; jcny[0]=1; *      for(LL i=1; i<=n+1; i++) jc[i]= (jc[i-1]*i)%Mod; $ ll x, y;Panax NotoginsengEXGCD (jc[n+1],mod,x,y); -jcny[n+1]= (x%mod+mod)%Mod; the      for(inti=n;i>=1; i--) jcny[i]= (jcny[i+1]* (i+1))%Mod; + } All C (intNintm) { the     return(((jc[n]*jcny[m)%mod) *jcny[n-m])%Mod; + } - intDfsintX1,intX2,intX3,intX4,intX5,intNumintRes) { $     intb[5]; $     intres=1; -     if(num==1){ -ans= (ans+res)%Mod; the         returnRes; -     }Wuyib[0]=x1;b[1]=x2;b[2]=x3;b[3]=x4;b[4]=X5; theStd::sort (b,b+num,cmp); -      for(intI=0; i<num;i++) Wu       for(intj=i+1; j<num;j++){ -B[I]+=B[J];inttmp=b[j];b[j]=0; Std::swap (b[j],b[num-1]); AboutRes+=dfs (b[0],b[1],b[2],b[3],b[4],num-1, (Res+b[i])%Mod); $Std::swap (b[j],b[num-1]); -b[i]-=tmp;b[j]=tmp; -     } -     returnRes;  A } + LL INV (ll a) { the ll x, y; - EXGCD (a,mod,x,y); $     returnx; the } the voidSbpianfen () { the     intK=dfs (a[1],a[2],a[3],a[4],a[5],n,0); theprintf"%d\n", ans);  - } inll A (intNintm) { the     return(Jc[n]*jcny[n-m])%Mod; the } About ll Pow (ll X,ll y) { thell res=1; the      while(y) { the         if(y%2) res= (res*x)%Mod; +x= (x*x)%Mod; -Y/=2; the     }Bayi     returnRes; the } the voidSxpianfen () { -ll res=0; -son[1]=1; son[0]=1; the      for(intI=2, num=n;i<=n;i++,num--) theSon[i]= (son[i-1]*c (NUM,2))%mod;//This line thenum[n-1]=1; num[n]=1; num[n+1]=1; num[n+2]=1; the      for(inti=n-2, num=2; i>=1; I--, num++) -Num[i]= (num[i+1]*c (num+1,2))%mod;//leaves the      for(intI=1, num=n;i<=n-1; i++,num--){ theRes= (res+ (son[i]*num[i+1])%mod* (num-1))%Mod; the     }94res= (res*sum)%Mod; theprintf"%lld\n", res); the } the intMain () {98n=read (); Init (); About      for(intI=1; i<=n;i++) -A[i]=read (), sum= (Sum+a[i])%Mod;101     if(n<=5) {Sbpianfen ();return 0;}102     if(n<=100000) {Sxpianfen ();return 0;}103}

Noi 10 even test DAY3 T1

Contact Us

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

  • Sales Support

    1 on 1 presale consultation

  • After-Sales Support

    24/7 Technical Support 6 Free Tickets per Quarter Faster Response

  • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.