Chocolate
| Time Limit: 2000MS | Memory Limit: 65536K | |||
| Total Submissions: 8800 | Accepted: 2306 | Special Judge | ||
Description
In 2100, ACM chocolate will be one of the favorite foods in the world.
"Green, orange, brown, red...", colorful sugar-coated shell maybe is
the most attractive feature of ACM chocolate. How many colors have you
ever seen? Nowadays, it‘s said that the ACM chooses from a palette of
twenty-four colors to paint their delicious candy bits.
One day, Sandy played a game on a big package of ACM chocolates
which contains five colors (green, orange, brown, red and yellow). Each
time he took one chocolate from the package and placed it on the table.
If there were two chocolates of the same color on the table, he ate both
of them. He found a quite interesting thing that in most of the time
there were always 2 or 3 chocolates on the table.
Now, here comes the problem, if there are C colors of ACM chocolates
in the package (colors are distributed evenly), after N chocolates are
taken from the package, what‘s the probability that there is exactly M
chocolates on the table? Would you please write a program to figure it
out?
Input
The input file for this problem contains several test cases, one per line.
For each case, there are three non-negative integers: C (C <= 100), N and M (N, M <= 1000000).
The input is terminated by a line containing a single zero.
Output
The output should be one real number per line, shows the probability for each case, round to three decimal places.
Sample Input
5 100 2 0
Sample Output
0.625
Source
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cstdlib>
5 #include<algorithm>
6 using namespace std;
7
8 int c,n,m;
9 double dp[2][10003];
10
11 int main()
12 {
13 // freopen("in.txt","r",stdin);
14 while(~scanf("%d",&c)){
15 if(c==0) break;
16 scanf("%d%d",&n,&m);
17 memset(dp,0,sizeof(dp));
18 if(m>n||m>c||(m+n)%2){//奇偶剪枝
19 printf("0.000\n");
20 continue;
21 }
22 if(n>1000)
23 {
24 n=1000+n%2;
25 }//1000以上都转化为1000或1001计算，因为只保留三位小数
26 dp[0][0]=1.0;dp[1][1]=1.0;
27 for(int i=2;i<=n;i++)
28 {
29 dp[i&1][0]=(dp[(i-1)&1][1])*1.0/c;
30 dp[i&1][c]=(dp[(i-1)&1][c-1])*1.0/c;
31 for(int j=1;j<=i&&j<=c;j++)
32 {
33 dp[i&1][j]=dp[(i-1)&1][j-1]*(c-j+1)*1.0/c+dp[(i-1)&1][j+1]*(j+1.0)/c;
34 }
35 }
36 printf("%.3lf\n",dp[n&1][m]);
37 }
38 return 0;
39 }