Coupons - UVa 10288 概率dp
2014-07-23 18:40
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Coupons
Input: standard input
Output: standard output
Time Limit: 2 seconds
Memory Limit: 32 MB
Coupons in cereal boxes are numbered 1 to n, and a set of one of each is required for a prize (a cereal box, of course). With one coupon per box, how many boxes on average are required to make a complete set of n coupons?
Input
Input consists of a sequence of lines each containing a single positive integer n, 1<=n<=33, giving the size of the set of coupons. Input is terminated by end of file.
Output
For each input line, output the average number of boxes required to collect the complete set of n coupons. If the answer is an integer number, output the number. If the answer is not integer, then output the integer part of the
answer followed by a space and then by the proper fraction in the format shown below. The fractional part should be irreducible. There should be no trailing spaces in any line of output.
题意:求每次抽都可能抽到1-n,问你集齐1-n的期望是多少。
思路:dp[i]表示在集齐i个的时候,到达目标还需要抽多少次的期望,dp
=0, dp[i]=i/n*dp[i]+(n-i)/i*dp[i+1],整理后发现ans=n+n/2+n/3+...+1。
打表代码如下:
AC代码如下:
Input: standard input
Output: standard output
Time Limit: 2 seconds
Memory Limit: 32 MB
Coupons in cereal boxes are numbered 1 to n, and a set of one of each is required for a prize (a cereal box, of course). With one coupon per box, how many boxes on average are required to make a complete set of n coupons?
Input
Input consists of a sequence of lines each containing a single positive integer n, 1<=n<=33, giving the size of the set of coupons. Input is terminated by end of file.
Output
For each input line, output the average number of boxes required to collect the complete set of n coupons. If the answer is an integer number, output the number. If the answer is not integer, then output the integer part of the
answer followed by a space and then by the proper fraction in the format shown below. The fractional part should be irreducible. There should be no trailing spaces in any line of output.
Sample Input
2
5
17
Sample Output
3
5
11 --
12
340463
58 ------
720720
题意:求每次抽都可能抽到1-n,问你集齐1-n的期望是多少。
思路:dp[i]表示在集齐i个的时候,到达目标还需要抽多少次的期望,dp
=0, dp[i]=i/n*dp[i]+(n-i)/i*dp[i+1],整理后发现ans=n+n/2+n/3+...+1。
打表代码如下:
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<cmath> typedef long long ll; using namespace std; ll gcd(ll a,ll b) { if(b==0) return a; else return gcd(b,a%b); } int main() { ll a=1,b=2,c,i,d,e,f,k; for(i=3;i<=33;i++) { a=a*i+b; b=b*i; c=gcd(a,b); a/=c; b/=c; d=a; e=b; d*=i; f=gcd(d,e); d/=f; e/=f; k=d%e; printf("n= %lld : %lld %lld %lld\n\n",i,d/e,d%e,e); } }
AC代码如下:
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
int main()
{ int n;
while(~scanf("%d",&n))
{ if(n==1)
printf("1\n");
else if(n==2)
printf("3\n");
else if(n==3)
{ printf(" 1\n");
printf("5 -\n");
printf(" 2\n");
}
else if(n==4)
{ printf(" 1\n");
printf("8 -\n");
printf(" 3\n");
}
else if(n==5)
{ printf("5\n");
printf("11 --\n");
printf("12\n");
}
else if(n==6)
{ printf(" 7\n");
printf("14 --\n");
printf(" 10\n");
}
else if(n==7)
{ printf(" 3\n");
printf("18 --\n");
printf(" 20\n");
}
else if(n==8)
{ printf(" 26\n");
printf("21 --\n");
printf(" 35\n");
}
else if(n==9)
{ printf("129\n");
printf("25 ---\n");
printf(" 280\n");
}
else if(n==10)
{ printf(" 73\n");
printf("29 ---\n");
printf(" 252\n");
}
else if(n==11)
{ printf("551\n");
printf("33----\n");
printf(" 2520\n");
}
else if(n==12)
{ printf("551\n");
printf("37 ----\n");
printf(" 2310\n");
}
else if(n==13)
{ printf(" 9473\n");
printf("41 -----\n");
printf(" 27720\n");
}
else if(n==14)
{ printf(" 13433\n");
printf("45 -----\n");
printf(" 25740\n");
}
else if(n==15)
{ printf(" 18581\n");
printf("49 -----\n");
printf(" 24024\n");
}
else if(n==16)
{ printf(" 4129\n");
printf("54 -----\n");
printf(" 45045\n");
}
else if(n==17)
{ printf("340463\n");
printf("58 ------\n");
printf("720720\n");
}
else if(n==18)
{ printf(" 620743\n");
printf("62 ------\n");
printf(" 680680\n");
}
else if(n==19)
{ printf(" 1662439\n");
printf("67 -------\n");
printf(" 4084080\n");
}
else if(n==20)
{ printf(" 3704479\n");
printf("71 -------\n");
printf(" 3879876\n");
}
else if(n==21)
{ printf(" 408335\n");
printf("76 ------\n");
printf(" 739024\n");
}
else if(n==22)
{ printf(" 46533\n");
printf("81 ------\n");
printf(" 235144\n");
}
else if(n==23)
{ printf(" 4597419\n");
printf("85 -------\n");
printf("5173168\n");
}
else if(n==24)
{ printf(" 9265735\n");
printf("90 --------\n");
printf(" 14872858\n");
}
else if(n==25)
{ printf(" 142406227\n");
printf("95 ---------\n");
printf(" 356948592\n");
}
else if(n==26)
{ printf(" 73762267\n");
printf("100 ---------\n");
printf(" 343219800\n");
}
else if(n==27)
{ printf(" 206234003\n");
printf("105 ----------\n");
printf(" 2974571600\n");
}
else if(n==28)
{ printf(" 2755866803\n");
printf("109 ----------\n");
printf(" 2868336900\n");
}
else if(n==29)
{ printf(" 71315126587\n");
printf("114 -----------\n");
printf(" 80313433200\n");
}
else if(n==30)
{ printf(" 65960897707\n");
printf("119 -----------\n");
printf(" 77636318760\n");
}
else if(n==31)
{ printf(" 1967151510157\n");
printf("124 -------------\n");
printf(" 2329089562800\n");
}
else if(n==32)
{ printf(" 3934303020314\n");
printf("129 -------------\n");
printf(" 4512611027925\n");
}
else if(n==33)
{ printf(" 4071048809039\n");
printf("134 -------------\n");
printf(" 4375865239200\n");
}
}
}
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