[Miscellaneous Questions] CSUOJ1276 Counting Route Sequence

http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1276

Test instructions: Go from point 1th to point N (each side can only walk once, the number of sides between the two nodes must be odd)

How many ways are there in different order of the nodes (e.g. 1->2->3->4 and 1->3->2->4 for two kinds)

Method number Modulo 1000000007

This problem only needs to consider the number of methods of alternating adjacent two nodes and then recursively multiply them.

is: If you walk from 1 to 5

Just consider the number of alternate methods of 2 and 3 first: (Obviously the number of combinations related to the number of edges)

Then a similar consideration of 3, 4 alternating method number

And then all of them are multiplied.

Formula is

The formula is

Because the range of N and M is 1e5, it is required to take the mold and therefore requires m! (N-M)! The inverse element

If the direct for once and then to TMP do EX_GCD really tle ...

1     LL tmp=1, ans=1; 2      for (LL i=min (n, m); i>=1; i--) 3     {4         tmp= (tmp*i)%mod; 5         Ans= (ans* (n+1-i))%mod; 6     }

So you can pre-preprocess the factorial in 1e5.

1#include <cstdio>2#include <cstdlib>3#include <cstring>4#include <climits>5#include <cctype>6#include <cmath>7#include <string>8#include <sstream>9#include <iostream>Ten#include <algorithm> One#include <iomanip> A using namespacestd; -#include <queue> -#include <stack> the#include <vector> -#include <deque> -#include <Set> -#include <map> +typedefLong LongLL; -typedefLong DoubleLD; + Const DoublePi=acos (-1.0); A Const Doubleeps=1e-9; at #defineINF 0x3f3f3f - #defineLson L, M, rt<<1 - #defineRson m+1, R, rt<<1|1 -typedef pair<int,int>PI; -typedef pair<int, PI >PP; - #ifdef _WIN32 in #defineLLD "%i64d" - #else to #defineLLD "%lld" + #endif - //#pragma COMMENT (linker, "/stack:1024000000,1024000000") the //ll Quick (ll A, ll b) {ll ans=1;while (b) {if (b & 1) ans*=a;a=a*a;b>>=1;} return ans;} * //inline int Read () {char ch= "; int ans=0;while (ch< ' 0 ' | | ch> ' 9 ') Ch=getchar (); while (ch<= ' 9 ' && Ch >= ' 0 ') {ans=ans*10+ch-' 0 '; Ch=getchar ();} return ans;} $InlinevoidPrint (LL x) {printf (LLD, x);p UTS ("");}Panax Notoginseng //inline void Read (LL &ret) {char c;int sgn; LL bit=0.1;if (C=getchar (), c==eof) return, while (c!= '-' &&c!= '. && (c< ' 0 ' | | C> ' 9 ')) C=getchar (), sgn= (c== '-'), -1:1;ret= (c== '-')? 0: (c ' 0 '); while (C=getchar (), c>= ' 0 ' &&c<= ' 9 ') ret=ret*10+ (c ' 0 '); if (c== "| | c== ' \ n ') {ret*=sgn; return;} while (C=getchar (), c>= ' 0 ' &&c<= ' 9 ') ret+= (c ' 0 ') *bit,bit/=10;ret*=sgn;} -  the Const intMod=1000000007; + inta[100005]; ALL jc[200005]; the voidPre () + { -jc[0]=1; $      for(intI=1; i<=200000; i++) $Jc[i]= (i*jc[i-1])%MoD; - } - voidEX_GCD (ll A, ll B, ll &x, LL &y) the { -     if(b)Wuyi     { theEX_GCD (b, a%b, x, y); -LL tmp=x; Wux=y; -Y=tmp-(A/b) *y; About     } $     Else -     { -x=1, y=0; -         return ; A     } + } the ll C (ll N, ll M) - { $     if(N==m | | m==0) the         return 1; the     if(m==1|| m==n-1) the         returnN; the //LL tmp=1, Ans=1; - //For (LL i=min (n, m); i>=1;i--) in //    { the //tmp= (tmp*i)%mod; the //ans= (ans* (n+1-i))%mod; About //    } the LL x, y; theEX_GCD (jc[m]*jc[n-m], mod, x, y); the     return(jc[n]*x)%MoD; + } - ll MOD (ll x) the {Bayi      while(x<0) thex+=MoD; the     returnX%MoD; - } - intMain () the { the pre (); the     intT; thescanf"%d", &t); -      while(t--) the     { the         intN; thescanf"%d", &n);94          for(intI=0; i<n-1; i++) thescanf"%d", &a[i]); theLL ans=1; the          for(intI=0; i<n-2; i++)98Ans= (Ans*c (a[i+1]-1)/2+ (a[i]-1)/2, (a[i]-1)/2)%mod)%MoD; About print (MOD (ans)); -     }101     return 0;102}

Csuoj 1276

[Miscellaneous Questions] CSUOJ1276 Counting Route Sequence