Bestcoder Round #35 (dzy Loves balls-violence DP)

Dzy Loves Ballsaccepts:371submissions:988Time limit:2000/1000 MS (java/others)Memory limit:65536/65536 K (java/others) Problem Description

There is n Black Balls and m White balls in the big box.

Now, Dzy starts-randomly pick out of the balls one by one. It forms a sequence S . If at the i -th operation, Dzy takes out of the black ball, S i =1 , otherwise S i =0 .

Dzy wants to know the expected times that ' "occurs in S .

Input

The input consists several test cases. ( T esTC ase≤ Max )

The first line contains the integers, n , m(1≤N,m≤ A)

Output

For each case, the output of the corresponding result, the format is P/q ( p and q is coprime)

Sample Input

1 12 3

Sample Output

1/26/5hintcase 1:s= ' or s= ', so the expected times = = = 1/2case 2:s= ' 00011 ' or s= ' 00101 ' or s= ' 00110 ' or s= ' 010 "or s= ' 01010 ' or s= ' 01100 ' or s= ' 10001 ' or s= ' 10010 ' or s= ' 10100 ' or s= ' 11000 ', so the expected times = (1+2+1+2+2+1+1+1 +1+0)/10 = 12/10 = 6/5

This time, I started DP-screwing.

Actually, there's a simpler way of knocking the formula, but I'm just going to hand in the table program.

Next_permunation This kind of violence is also possible??

#include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> #include < functional> #include <iostream> #include <cmath> #include <cctype> #include <ctime>using namespace std; #define for (I,n) for (int. i=1;i<=n;i++) #define FORK (I,k,n) for (int. i=k;i<=n;i++) #define REP (I,n) for (int i=0;i<n;i++) #define ForD (I,n) for (int. i=n;i;i--) #define REPD (I,n) for (int. i=n;i>=0;i--) #define FORP (x) for ( int p=pre[x];p; p=next[p]) #define FORPITER (x) for (int &p=iter[x];p; p=next[p]) #define LSON (x<<1) #define Rson ((x<<1) +1) #define MEM (a) memset (A,0,sizeof (a)), #define MEMI (a) memset (A,127,sizeof (a)), #define MEMI (a) memset ( A,128,sizeof (a)); #define INF (2139062143) #define F (100000007) #define MAXN (+) typedef long Long Ll;ll Mul (ll a,ll b) {RE Turn (a*b)%F;} ll Add (ll A,ll b) {return (a+b)%F;} ll Sub (ll A,ll b) {return (a-b+ (a)/f*f+f)%F; void Upd (ll &a,ll b) {a= (a%f+b%f)%F;} ll GCD (ll A,ll b) {if (b==0) return A;return gcd (b,a%b);} ClThe Fenshu{public:ll A, B,//denominator a molecule B. Fenshu (ll _b,ll _a): A (_a), B (_b) {}fenshu (): A (1), B (0) {}void relax () {int t=gcd (a, b); =t,b/=t;} Friend Fenshu operator* (Fenshu A,fenshu b) {Fenshu C=fenshu (A.B*B.B,A.A*B.A); C.relax (); return c;} Friend Fenshu operator+ (Fenshu A,fenshu b) {Fenshu C=fenshu (A.A*B.B+A.B*B.A,A.A*B.A); C.relax (); return c;} void pri () {cout<<b<< '/' <<a;}} F[maxn][maxn][2];int n,m;void turn_out (int n,int m) {Fenshu T=f[n][m][0]*fenshu (m,n) +f[n][m][1]*fenshu (n-m,n); T.pri ( );} int main () {//freopen ("a.in", "R", stdin); n=24,m=12; Fork (i,2,n) Rep (j,i+1) {int K=0;f[i][j][k]=f[i-1][j-1][0]*fenshu (j-1,i-1) +f[i-1][j-1][1]*fenshu (i-j,i-1); K=1;f[i] [j] [K]= (Fenshu () +f[i-1][j][0]) *fenshu (j,i-1) +f[i-1][j][1]*fenshu (i-1-j,i-1);} /*for (i,24) {Rep (J,min (i+1,13)) {turn_out (i,j);cout<< ';} Cout<<endl;} */while (scanf ("%d%d", &n,&m) ==2) {N+=m;fenshu T=f[n][m][0]*fenshu (m,n) +f[n][m][1]*fenshu (n-m,n); T.pri (); Cout<<endl;} return 0;}

Bestcoder Round #35 (dzy Loves balls-violence DP)