visit (Lucas Theorem && Chinese remainder theorem)

So obviously the number of combinations I treat it as a DP do I really. . . .

Because the start and end points have been identified, we found that if we identified a number of steps in the other direction will determine the direction

Combination of several approaches 1:

A step is provided right away, then go left of b = an step, set up is c, down to D;

c+d=t-a-b; c-d=m;

求出c=（t+n+m-i-i）/2;if(c%2)continue;

(Because if c is not divisible by the number of steps to the right to walk two said it could not go back)

Combination of several approaches 2:

Reference nc God Ben practice

The horizontal direction is provided first go i-step total, so (in) / 2 to return the number of steps in the horizontal direction,

The number of steps in the vertical direction ti, is returned in the same way (tim) / 2;

式子C(t,i)*C(i,(i-n)/2)*C(t-i,(t-i-m)/2)

(Vertical + horizontal direction since the number of steps = t, it is not by C (ti, ti))

As Lucas and Chinese Remainder Theorem

We found that the data is more the product of prime numbers, apparently can be solved with the Chinese remainder theorem, of course, to advance the quality factor decomposition

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<string>
  5 #include<algorithm>
  6 #include<cmath>
  7 #include<stack>
  8 #include<vector>
  9 #include<queue>
 10 #define MAXN 100001
 11 #define ps push_back
 12 #define ll long long
 13 using namespace std;
 14 vector<int>su;ll jie[MAXN];
 15 void fen(int x)
 16 {
 17    for(int i=2;i<=sqrt(x);++i)
 18    {
 19        if(x%i==0)
 20        {
 21            while(x%i==0)
 22            {
 23                x/=i;
 24            }
 25            su.ps(i);
 26        }
 27    }
 28    if(x!=1)
 29    {
 30           su.ps(x);
 31    }
 32 }
 33 ll pow(ll x,ll y,ll mod)
 34 {
 35     ll ans=1;
 36     if(y==0)return 1;
 37     while(y)
 38     {
 39        if(y&1)ans=ans*x%mod;
 40        x=x*x%mod;
 41        y>>=1;
 42     }
 43     return ans%mod;
 44 }
 45 ll C(ll x,ll y,ll mod)
 46 {
 47     if(y>x)return 0;
 48     if(y==0)return 1;
 49     return jie[x]*pow(jie[y]*jie[x-y]%mod,mod-2,mod)%mod;
 50 }
 51 ll lus(ll x,ll y,ll mod)
 52 {
 53    if(y>x)return 0;
 54    if(y==0)return 1;
 55    return lus(x/mod,y/mod,mod)*C(x%mod,y%mod,mod)%mod;
 56 }
 57 ll M[MAXN],ans[MAXN];
 58 void exgcd(ll a,ll b,ll &x,ll &y)
 59 {
 60    if(b==0){
 61       x=1;y=0;return ;
 62    }
 63    exgcd(b,a%b,x,y);
 64    ll z=x;x=y;y=z-(a/b)*y;
 65    return ;
 66 }
 67 ll sum;ll len=1;
 68 void CRT()
 69 {
 70    for(ll i=0;i<su.size();++i)
 71    {
 72       len*=su[i];    
 73       //printf("%lld\n",len);  
 74    }
 75    for(ll i=0;i<su.size();++i)
 76    {
 77       M[i]=len/su[i];
 78    }
 79    for(ll i=0;i<su.size();++i)
 80    {
 81         ll x,y;
 82         exgcd(M[i],su[i],x,y);
 83         x=(x+su[i])%su[i];
 84         sum=(sum+ans[i]*M[i]*x)%len;
 85         //printf("sum=%lld ans=%lld M=%lld x=%lld\n",sum,ans[i],M[i],x);
 86    }
 87    printf("%lld\n",sum);
 88 }
 89 ll t;
 90 void fir(ll mod)
 91 {
 92     for(ll i=1;i<=min(mod,t);++i)
 93     {
 94         jie[i]=(jie[i-1]*i)%mod;
 95     }
 96 }
 97 ll MOD,n,m;
 98 int main()
 99 {
100       scanf("%lld%lld",&t,&MOD);
101       scanf("%lld%lld",&n,&m);
102       if(n<0)n=-n;
103       if(m<0)m=-m;
104       jie[0]=1;    
105       fen(MOD);
106       for(ll k=0;k<su.size();++k)
107       {
108           fir(su[k]);
109           ll mod=su[k];
110           //printf("mod=%lld\n",mod);
111           for(ll i=n;i<=t-m;++i)
112           {     
113             ll a=i;
114             ll b=i-n;
115             ll c=(t+n+m-i-i);
116             if(c%2!=0)continue;c/=2ll;
117             ll d=t-a-b-c;
118             ans[k]=(ans[k]+lus(t,a,mod)*lus(t-a,b,mod)%mod*lus(t-a-b,c,mod)%mod+mod)%mod;
119           }
120      }
121      CRT();
122 }

View Code