HDU 5852 Intersection is not allowed!

Problem Description

There are K pieces on the chessboard.

The size of the chessboard is N*N.

The pieces are initially placed on the top cells of the board.

A piece located on (r, c) can be moved by one cell right to (r, c + 1) or one cell down to (r+1, c).

Your task is to count how many different ways to move all pieces to the given positions at the bottom of the board.

Furthermore, the paths of the pieces mustn’t intersect each other.

 

Input

The first line of input contains an integer T-the number of test cases.

Each test case begins with a line containing two integers-N(1<=N<=100000) and K(1<=K<=100) representing the size of the chessboard and the number of pieces respectively.

The second line contains K integers: 1<=a1<a2< …<aK<=N representing the initial positions of the pieces. That is, the pieces are located at (1, a1), (1, a2), …, (1, aK).

Next line contains K integers: 1<=b1<b2<…<bK<=N representing the final positions of the pieces. This means the pieces should be moved to (N, b1), (N, b2), …, (N, bK).

 

Output

Print consecutive T lines, each of which represents the number of different ways modulo 1000000007.

 

Sample Input

15 21 23 4

 

Sample Output

50

 

Author

金策工业综合大学（DPRK）

 

Source

2016 Multi-University Training Contest 9

 

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组合数问题+行列式求值+逆元～

思路见：http://blog.csdn.net/v5zsq/article/details/52419187

建议计算过程开long long，容易写。

组合数一定要先判掉m>n的情况，直接输出0！

#include<cstdio>#include<iostream>using namespace std;#define ll long long#define mod 1000000007ll t,n,m,a[101][102],b[101],c[101],sheng[200001],ni[200001],ans;ll read(){ll x=0,f=1;char ch=getchar();while(ch<'0' || ch>'9') {if(ch=='-') f=-1;ch=getchar();}while(ch>='0' && ch<='9') {x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}return x*f;}ll mi(ll u,ll v){ll now=1;u%=mod;for(;v;v>>=1,u=(ll)u*u%mod) if(v&1) now=(ll)now*u%mod;return now;}ll cc(ll n,ll m){if(m>n) return 0;return (ll)sheng[n]*ni[m]%mod*ni[n-m]%mod;}ll cal(ll n){ll ans=1,pos=-1;for(ll i=1;i<=n;i++){pos=-1;for(ll j=i;j<=n;j++) if(a[j][i]){pos=j;break;}if(pos==-1) return 0;if(pos!=i) for(ll j=i;j<=n;j++) swap(a[pos][j],a[i][j]);ll now=mi(a[i][i],mod-2);for(ll j=i+1;j<=n;j++)  if(a[j][i])  {  ans=ans*now%mod;  for(ll k=i+1;k<=n;k++) a[j][k]=((a[j][k]*a[i][i]%mod-a[i][k]*a[j][i]%mod)+mod)%mod;  a[j][i]=0;  }}for(ll i=1;i<=n;i++) ans=ans*a[i][i]%mod;return ans;}int main(){t=read();sheng[0]=1;ni[0]=1;for(int i=1;i<=200000;i++) sheng[i]=(ll)sheng[i-1]*i%mod,ni[i]=mi(sheng[i],mod-2);while(t--){n=read();m=read();for(int i=1;i<=m;i++) b[i]=read();for(int i=1;i<=m;i++) c[i]=read();for(int i=1;i<=m;i++)  for(int j=1;j<=m;j++) a[i][j]=cc(n-1-b[j]+c[i],n-1);ans=cal(m);printf("%lld\n",ans);}}