CodeForces 48C D

每个加油的站可以确定一个alpha的上下界，比如，第i次加油站a[i]，前面加了i次油，a[i]*10≤ alpha*i <(a[i]+1)*10。

取最大的下界，取最小的上界，看看两者之间的满足条件的下一个加油站是否唯一。

因为可以用分数，所有就没用double了

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a; }
struct Fra
{
    ll p,q;
    Fra(ll x = 0,ll y = 1):p(x),q(y){ normal(p,q); }
    void normal(ll &p,ll &q) { ll g = gcd(p,q); p/=g; q/=g; }
    Fra operator = (int x) {  p = x; q = 1; return *this; }
    Fra operator = (ll x) { p = x; q = 1; return *this; }
    Fra operator - () { return {-p,q}; }
    Fra operator + (Fra &r) {
        ll m,n;
        m = p*r.q+r.p*q;
        n = q*r.q;
        normal(m,n);
        return {m,n};
    }
    Fra operator += (Fra& r) { return *this = *this+r; }
    Fra operator - (Fra &r) { return (-r) + *this; }
    Fra operator -= (Fra &r) { return *this = *this-r; }
    Fra operator * (Fra &r) {
        ll m,n;
        m = p*r.p;
        n = q*r.q;
        normal(m,n);
        return {m,n};
    }
    Fra operator *= (Fra &r) { return (*this) = (*this)*r; }
    Fra operator /(Fra &r) { return Fra(r.q,r.p) * (*this); }
    Fra operator /=(Fra &r) { return (*this) = (*this)/r; }
    bool operator == (const Fra& r) const { return p*r.q == r.p*q; }
    bool operator < (const Fra& r) const { return  p*r.q < r.p*q; }
    void print() { normal(p,q); if(q<0)q = -q,p = -p; printf("%lld/%lld
",p,q); }
};

const int maxn = 1e3+5;

const ll INF = 1e16;
int main()
{
    //freopen("in.txt","r",stdin);
    int n; scanf("%d",&n);
    Fra Low(0),High(INF);
    for(int i = 1; i <= n; i++){
        int t; scanf("%d",&t);
        Low = max(Fra(t*10,i),Low);
        High = min(Fra(t*10+10,i),High);
    }
    Low = Fra(n+1)*Low;
    High = Fra(n+1)*High;
    int u = (High.p-1)/High.q;
    int d = (Low.p)/Low.q;
    if(u/10 != d/10) {
        puts("not unique");
    }else {
        printf("unique
%d",d/10);
    }
    return 0;
}