UVA 10288 - Coupons
题意：n个张票，每张票取到概率等价，问连续取一定次数后，拥有全部的票的期望
思路：递推。f[i]表示还差i张票的时候期望，那么递推式为
f(i)=f(i)∗(n−i)/n+f(i−1)∗i/n+1
化简后递推就可以，输出要输出分数比較麻烦
代码：
#include <cstdio>
#include <cstring>
#include <cmath> long long gcd(long long a, long long b) {
if (!b) return a;
return gcd(b, a % b);
} long long lcm(long long a, long long b) {
a = a / gcd(a, b) * b;
if (a < 0) a = -a;
return a;
} struct Fraction {
long long a, b;
Fraction() {a = 0; b = 1;} Fraction(long long x) {
a = x; b = 1;
} Fraction(long long x, long long y) {
a = x; b = y;
} void deal() {
if (b < 0) {b = -b; a = -a;}
long long k = gcd(a, b);
if (k < 0) k = -k;
a /= k; b /= k;
} Fraction operator+(Fraction p) {
Fraction ans;
ans.b = lcm(b, p.b);
ans.a = ans.b / b * a + ans.b / p.b * p.a;
ans.deal();
return ans;
} Fraction operator-(Fraction p) {
Fraction ans;
ans.b = lcm(b, p.b);
ans.a = ans.b / b * a - ans.b / p.b * p.a;
ans.deal();
return ans;
} Fraction operator*(Fraction p) {
Fraction ans;
ans.a = a * p.a;
ans.b = b * p.b;
ans.deal();
return ans;
} Fraction operator/(Fraction p) {
Fraction ans;
ans.a = a * p.b;
ans.b = b * p.a;
ans.deal();
return ans;
} void operator=(int x) {
a = x;
b = 1;
} void print() {
if (a == 0) {printf("0\n"); return;}
if (a % b == 0) {printf("%lld\n", a / b); return;}
int sn = 0;
if (a / b > 0) {
long long num = a / b;
while (num) {
num /= 10;
sn++;
}
}
if (sn) for (int i = 0; i <= sn; i++) printf(" ");
printf("%lld\n", a % b);
long long num = b;
int cnt = 0;
while (num) {
num /= 10;
cnt++;
}
printf("%lld ", a / b);
for (int i = 0; i < cnt; i++) printf("-");
printf("\n");
for (int i = 0; i <= sn; i++) printf(" ");
printf("%lld\n", b);
}
}; Fraction dp[35][35];
int n; int main() {
for (long long i = 1; i <= 33; i++) {
dp[i][0] = 0;
for (long long j = 1; j <= i; j++)
dp[i][j] = (dp[i][j - 1] * Fraction(j, i) + Fraction(1)) * Fraction(i, j);
}
while (~scanf("%d", &n)) {
dp[n][n].print();
}
return 0;
}