uva 10288 gailv
阿新 • • 發佈:2018-08-26
ios 我們 else with sca def for each != code
Problem F 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 setofn coupons? Input Input consists of a sequence of lines each containing a single positive integern, 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 completeset ofn 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 inany line of output. Sample Input 2 5 17 Sample Output 3 5 11 -- 12 340463 58 ------
記得每次求gcd不然會爆掉。 思路:到k張時。我們還有n-k張沒有得到。所以我們得到的概率為n-k/n;
所以到k+1步的期望時n/n-k;
整理得 n* 1/i(n>i>1)的累加
#include <iostream> #include <string.h> #include <stdio.h> #include <algorithm> using namespace std; #define ll long long ll gcd(ll n,ll m) { if(m==0) return n; return gcd(m,n%m); } int geth(ll x) { int ans=0; while(x>0) { x=x/10; ans++; } return ans; } int main() { ll n; while(~scanf("%lld",&n)) { ll fenmu=1; ll temp=1; for(int i=1;i<=n;i++) { temp=gcd(fenmu,i); fenmu=fenmu*i/temp; } ll fenzi=0; for(int i=1;i<=n;i++) { fenzi+=fenmu/i; } fenzi=fenzi*n; ll p=gcd(fenzi,fenmu); fenzi/=p; fenmu/=p; ll t=fenzi/fenmu; if(fenzi-fenmu*t>0) { if(t!=0) { ll a=geth(t); for(int i=0;i<=a;i++) {cout<<" ";} cout<<fenzi-fenmu*t; ll b=geth(fenzi-fenmu*t); ll c=geth(fenmu); if(b>c) c=b; cout<<endl; cout<<t<<" "; for(int i=1;i<=c;i++) {cout<<"-";} cout<<endl; for(int i=0;i<=a;i++) {cout<<" ";} cout<<fenmu<<endl; } } else cout<<t<<endl; } return 0; }
uva 10288 gailv