[Number of combinations] [High Precision] Jzoj P3431 network

Description

A city street grid-like form, the lower left corner coordinates A (0, 0), the coordinates of the upper right corner of B (n, m), where n> = m. Now from A (0, 0) point of view, or only just above the front-right to walk along the street, and not a straight line through the points shown in the upper left, i.e., any route of the point (x, y) must meet x> = y, Will these premises, arrival B (n, m) how many moves.

 

answer

• From (0,0) to (n, m) is the number of random walk scenario C (n + m, m)
• Then the subject requirements can not pass this line y = x, i.e. not touch y = x + 1 is linear
• We will then consider an illegal route along the y = x + 1 symmetric elapsed, (n, m) is symmetrical point in the past (m-1, n + 1)
• Consider the proof number Catalan, we found go from the origin point of symmetry (m-1, n + 1 ) can be a path symmetric back and comes to an end corresponding to one from the origin (n, m) passing through the y = x the straight-line path
• Therefore, through the y = x is the number of paths from the origin to the end of the path is symmetrical i.e. C (n + 1 + m-1, m-1) = C (n + m, m-1)
• All the answer is C (n + m, m) = C (n + m, m-1)
• The formula can simplify it:
• 
• The common denominator was:
• 
• Finally, make a precision on the ok

 

Code

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 using namespace std;
 5 const int N=10000,M=N*10;
 6 struct num
 7 {
 8     int shu[N];
 9     friend num operator-(num y,num z) 
10     {
11         num x;memset(x.shu,0,sizeof(x.shu)),x.shu[0]=y.shu[0];
12         for (int i=1;i<=x.shu[0];i++) 
13         {
14             x.shu[i]+=y.shu[i]-z.shu[i];
15             if (x.shu[i]<0) x.shu[i]+=M,x.shu[i+1]--;
16         }
17         while (!x.shu[x.shu[0]]) x.shu[0]--;
18         return x;
19     }
20     friend num operator*(num y,int z) 
21     {
22         num x;memset(x.shu,0,sizeof(x.shu)),x.shu[0]=y.shu[0];
23         for (int i=1;i<=x.shu[0];i++) x.shu[i]+=y.shu[i]*z,x.shu[i+1]+=x.shu[i]/M,x.shu[i]%=M;
24         while (x.shu[x.shu[0]+1]) x.shu[0]++;
25         return x;
26     }
27 }ans;
28 int p[N+5],c[N+5],n,m;
29 void recout(int x,int f) { for(;x!=1;x/=p[x]) c[p[x]]+=f; }
30 num C(int m,int n) 
31 {
32     num x;memset(x.shu,0,sizeof(x.shu)),x.shu[0]=x.shu[1]=1,memset(c,0,sizeof(c));
33     for (int i=n+1;i<=m;i++) recout(i,1);
34     for (int i=1;i<=m-n;i++) recout(i,-1);
35     for (int i=1;i<=N;i++) for (int j=1;j<=c[i];j++) x=x*i;
36     return x;
37 }
38 int main() 
39 {
40     for (int i=2;i<=N;i++) if (!p[i]) for (int j=1;j<=N/i;j++) p[i*j]=i;
41     scanf("%d%d",&n,&m),ans=C(m+n,m)-C(m+n,m-1),printf("%d",ans.shu[ans.shu[0]]);
42     for (int i=ans.shu[0]-1;i>=1;i--) printf("%05d",ans.shu[i]);
43 }