hdoj 1099 Lottery（期望）

【题目大意】：给出n张不同的牌，每种无限张，问你收集齐n张不同的牌所需的平均次数。

【解题思路】：简单的期望题，假设现在手中有k张不同的牌，则下一张是不同的牌的概率是 (n-k)/n。那么抽到不重复的牌的期望张数是 n/(n-k)...

                            把n提出来~求E=n*sigema(1/k)（1<=k<=n）

                            一时大意，求分母的时候没边求边约分，爆longlong了wa了一次，没意识啊。

【代码】：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <queue>#include <cmath>#include <string>#include <cctype>#include <map>#include <iomanip>                   using namespace std;                   #define eps 1e-8#define pi acos(-1.0)#define inf 1<<30#define linf 1LL<<60#define pb push_back#define lc(x) (x << 1)#define rc(x) (x << 1 | 1)#define lowbit(x) (x & (-x))#define ll long longint n;ll get_len(ll n){    ll len=0;    while (n){        n/=10,len++;    }    return len;}ll GCD(ll a,ll b){    while (b) {        ll c;        c=a%b,a=b,b=c;     }    return a;}void solve() {    ll tmp=1;    for (int i=1; i<=n; i++) tmp=tmp*i/GCD(tmp,i);    ll ans=0;    for (int i=1; i<=n; i++) ans+=tmp/i;    ans*=n;    ll gcdtmp;    gcdtmp=GCD(ans,tmp);    ans=ans/gcdtmp;    tmp=tmp/gcdtmp;    ll k;    k=ans/tmp;    ans=ans%tmp;    if (ans==0) {cout << k << endl;}    else {        ll len1,len2;        len1=get_len(k);        len2=get_len(tmp);        for (int i=1; i<=len1+1; i++) cout <<" " ;        cout << ans << endl;        cout << k << " ";        for (int i=1; i<=len2; i++) cout << "-";        cout << endl;        for (int i=1; i<=len1+1; i++) cout << " ";        cout << tmp << endl;    }}int main() {    while (~scanf("%d",&n)){        solve();    }    return 0;}