UVA10288 Coupons

题意：n个物品(兑换券)，每次从中选择一个(不取出)，求期望多少次可以取遍所有的物品n<=33。

题解：先说一下概率吧：显然为1 * (n-1)/n * (n-2)/n * ... * 1/n 。这道题要求的期望一定不能直接取概率的倒数，因为直接取倒数的意义是一次取n个，期望多少次可以取遍，与题目不符。应该对于每一步取倒数然后相加作为答案。ans = 1 * n/(n-1) * n/(n-2) ... * n。实现的时候用long long，并且注意要不停地约分，不然会爆long long。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define LL long long
LL n;
LL fz, fm; //当前分子和分母
LL adz, adm; //增加的分子和分母
LL gcd(LL a, LL b) {
if (!b) return a;
return gcd(b, a%b);
}
void add() {
LL g = gcd(fm, adm);
LL t1 = fz*adm/g+adz*fm/g, t2 = fm/g*adm;
fz = t1; fm = t2;
g = gcd(fz, fm);
fz/=g; fm/=g;
}
int lg(LL n)
{
int cnt = 0;
while (n) ++cnt, n/=10;
return cnt;
}
void put(int cnt, char c) {
while (cnt>0) putchar(c), --cnt;
}
void print() {
if (fz % fm == 0) {
cout << fz << '\n';
return;
}
LL a = fz/fm , b = fz%fm, c = fm;
put(lg(a) + 1, ' ');
cout << b << '\n';
cout << a << ' ';
put(lg(c), '-');
put(1, '\n');
put(lg(a) + 1, ' ');
cout << c << '\n';
}
int main()
{
while (cin >> n) {
LL i;
fz = 0; fm = 1;
adz = 0; adm = 1;
for (i = 0; i<n; ++i) {
adz = n; adm = n - i;
add();
}
print();
}
return 0;
}