題目鏈接

luogu P1291 [SHOI2002]百事世界杯之旅

題解

設\(f[k]\)表示還有\(k\)個球員沒有收集到的概率
再買一瓶，買到的概率是\(k/n\)，買不到的概率是\((n-k) /k\)
那麽\(f[k] = f[k]*(n-k)/n + f[k-1]*k/n + 1\)
移向一下\(f[k] = f[k-1] + n/k\)

代碼

#include<cstdio>
#include<cstring>
#include<algorithm>

inline int read()  {
    int x = 0,f = 1;
    char c = getchar();
    while
 
(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
    while(c <= '9' && c >='0') x = x * 10 + c - '0',c = getchar();
    return x*f;
}
#define LL long long
LL gcd(LL x,LL y) {return y == 0 ? x : gcd(y,x % y);}
int n;
int main() {
    n = read();
    LL fz = n, fm = 1
 
,Tfz,Tfm;
    for(int i = 2;i <= n;++ i) {
        Tfz = n,Tfm = i;
        LL _gcd = gcd(Tfm,fm);
        fz = fz * (Tfm / _gcd) + Tfz * (fm / _gcd);
        fm *= (Tfm/_gcd);
        _gcd = gcd(fz,fm);
        fz /= _gcd,fm /= _gcd;
    }
    if(fm == 1) {printf("%lld",fz);return 0;}
    LL x = fz / fm;fz %= fm;
    LL tx = x,cnt=0
 
;
    while(tx) cnt ++,tx /= 10;
    for(int i = 1;i <= cnt;++ i) printf(" ");
    printf("%lld\n",fz);
    if(x) printf("%lld",x);
    tx = fm;
    while(tx) printf("-"),tx /= 10; printf("\n");
    for(int i = 1;i <= cnt;++ i) printf(" ");
    printf("%lld",fm);
    return 0;
}

luogu P1291 [SHOI2002]百事世界杯之旅