BestCoder Round #35(DZY Loves Balls-暴力dp)

DZY Loves Balls

Accepts: 371

Submissions: 988

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 65536/65536 K (Java/Others)

Problem Description

There are n  black balls and m  white balls in the big box.

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

DZY wants to know the expected times that '01' occurs in S .

Input

The input consists several test cases. (TestCase≤150 )

The first line contains two integers, n ,m(1≤n,m≤12) 

Output

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

Sample Input

1 12 3

Sample Output

1/26/5
Hint
Case 1: S='01' or S='10', so the expected times = 1/2 = 1/2Case 2: S='00011' or S='00101' or S='00110' or S='01001' 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

这次学乖，开始DP乱搞

其实还有更简单的敲公式法，，不过我就直接把打表程序交了

next_permunation 这种暴力也是可以的？？

#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 (100) typedef long long ll;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+(a-b)/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);}class fenshu{public:ll a,b; //分母a 分子b fenshu(ll _b,ll _a):a(_a),b(_b){}fenshu():a(1),b(0){}void relax(){int t=gcd(a,b);a/=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(1,1)+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;}