2019牛客多校-Day1

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B. Integration

题意

已知 $\Large \int_{0}^{\infty} \frac{1}{1+x^{2}} \mathrm{d} x=\frac{\pi}{2}$ ，求 $\Large\frac{1}{\pi} \int_{0}^{\infty} \frac{1}{\prod_{i=1}^{n}\left(a_{i}^{2}+x^{2}\right)} \mathrm{d} x$

题解

由 $\large\frac 1 a\times \frac 1 b=(\frac 1 a-\frac 1 b)\times \frac 1 {(b-a)}$ ，获得启发，化乘为加，令 $\large c_i=\prod\limits_{j\ne i}\frac 1 {a_j^2-a_i^2}$ ，则 $\large\prod \frac 1 {(a_i^2+x^2)}=\sum \frac {c_i}{a_i^2+x^2}$

代码

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#include <bits/stdc++.h>
using namespace std;
#define forl(i, l, r) for (int i = l; i <= r; i++)
#define forr(i, r, l) for (int i = r; i >= l; i--)
#define for1(i, n) for (int i = 1; i <= n; i++)
#define fro1(i, n) for (int i = 1; i <= n; i++)
#define for0(i, n) for (int i = 0; i < n; i++)
#define fro0(i, n) for (int i = 0; i < n; i++)
#define meminf(a) memset(a, inf, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define mem0(a) memset(a, 0, sizeof(a))
#define memcp(a,b) memcpy(a,b,sizeof(b))
#define oper(type) bool operator <(const type y)const
#define mp make_pair
#define pu_b push_back
#define pu_f push_front
#define po_b pop_back
#define po_f pop_front
#define fi first
#define se second
#define whiel while
#define retrun return
typedef pair<long long, long long> pll;
typedef vector<long long> vll;
typedef pair<int, int> pii;
typedef unsigned long long ull;
typedef vector<int> vii;
typedef double db;
typedef long long ll;
typedef int itn;
void in(initializer_list<int*> li){for(auto ptr=li.begin();ptr!=li.end();ptr++)scanf("%d",*ptr);}
void in(initializer_list<ll*> li){for(auto ptr=li.begin();ptr!=li.end();ptr++)scanf("%lld",*ptr);}
void in(initializer_list<db*> li){for(auto ptr=li.begin();ptr!=li.end();ptr++)scanf("%lf",*ptr);}
void out(initializer_list<int> li){for(auto ptr=li.begin();ptr!=li.end();ptr++)printf("%d ",*ptr);}
void outln(initializer_list<int> li){for(auto ptr=li.begin();ptr!=li.end();ptr++)printf("%d\n",*ptr);}
void out(initializer_list<ll> li){for(auto ptr=li.begin();ptr!=li.end();ptr++)printf("%lld ",*ptr);}
void outln(initializer_list<ll> li){for(auto ptr=li.begin();ptr!=li.end();ptr++)printf("%lld\n",*ptr);}
int in(int &a,int &b,int &c,int &d){return scanf("%d%d%d%d",&a,&b,&c,&d);}
int in(int &a,int &b,int &c){return scanf("%d%d%d",&a,&b,&c);}
int in(int &a,int &b){return scanf("%d%d",&a,&b);}
int in(ll &a){return scanf("%lld",&a);}
int in(int &a){return scanf("%d",&a);}
int in(char *s){return scanf("%s",s);}
int in(char &c){return scanf("%c",&c);}
void out(int a,bool ln){printf("%d%c",a,ln? '\n':' ');}
void out(ll a,bool ln){printf("%lld%c",a,ln? '\n':' ');}
void out(db a,int digit,bool ln){printf("%.*f%c",digit,a,ln? '\n':' ');}
const db pi = acos((db)-1);
const ll inf =0x3f3f3f3f;
const db eps = 1e-8;
const int N = 1.1e3;
const int M = 2.1e5;
const ll mod = 1e9+7;
int sign(db a) {return a < -eps ? -1 : a > eps;}
int db_cmp(db a, db b){ return sign(a-b);}

int a[N];
ll qPow(ll a, ll b, ll c) {  //求(a^b) % c
    ll ret = 1;
    while (b) {
        if (b & 0x1) ret = ret * a % c;
        a = a * a % c;
        b >>= 1;
    }
    return ret;
}
int main() {
#ifdef PerpEternal
    freopen("/Users/perpeternal/Documents/Sketch/data/in.dat", "r", stdin);
    // freopen("/Users/perpeternal/Documents/Sketch/data/out.dat", "w", stdout);
#endif
    //ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int n;
    while(~in(n)){
        for1(i,n)in(a[i]);
        // if(n==1){
        //     out(qPow(2ll*a[1],mod-2,mod),1);
        //     continue;
        // }
        ll ans=0;
        for1(i,n){
            ll tm=1;
            for1(j,n){
                if(j!=i){
                    tm=tm*((1ll*a[j]*a[j]%mod-1ll*a[i]*a[i]%mod+mod)%mod)%mod;
                    // tm=tm*qPow((1ll*a[j]*a[j]%mod-1ll*a[i]*a[i]%mod+mod)%mod,mod-2,mod)%mod;
                }
            }
            ans=(ans+qPow(2*a[i],mod-2,mod)*qPow(tm,mod-2,mod)%mod)%mod;
        }
        out(ans,1);
    }
    return 0;
}

C. Euclidean Distance

题意

给一个点 $(\frac {a_1} m,\frac {a_2} m,\cdots,\frac {a_n} m)(-m\le a_i\le m)$ ，找到一个点 $P(p_1,p_2,\cdots,p_n)(p_i\ge 0,\sum p_i=1)$ 使得 $\sum (\frac {a_i}m-p_i)^2$ 最小

题解

对 $a_i$ 排序，逐渐减小最大的 $a_i$ ，使得序列的最大值尽可能小

代码

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bool cmp(int x,int y){
    return x>y;
}
ll gcd(ll x,ll y){
    return y?gcd(y,x%y):x;
}
struct fraction{
    ll son, mom;
    fraction(){}
    fraction(ll a,ll b){
        if(a){ll gc=gcd(abs(a),abs(b));if(b<0) son=-a/gc,mom=-b/gc;else son=a/gc,mom=b/gc;}
        else son=0,mom=1;
    }
    bool operator==(const fraction &x)const{
        return son==x.son&&mom==x.mom;
    }
    bool operator!=(const fraction &x)const{
        return !(son==x.son&&mom==x.mom);
    }
    bool operator<(const fraction &y)const{
        return son * y.mom < y.son * mom;
    }
    bool operator<=(const fraction &y)const{
        return son * y.mom <= y.son * mom;
    }
    fraction operator*(const fraction &y)const{
        return fraction(son*y.son,mom*y.mom);
    }
    fraction operator+(const fraction &y)const{
        return fraction(son * y.mom + y.son * mom,mom*y.mom);
    }
    fraction operator/(const fraction &y)const{
        return fraction(son*y.mom,mom*y.son);
    }
    fraction operator-(const fraction &y)const{
        return fraction(son * y.mom - y.son * mom,mom*y.mom);
    }
    void display(){
        if(mom==1)printf("%lld\n",son);
        else printf("%lld/%lld\n",son,mom);
    }
};
int main() {
#ifdef PerpEternal
    // freopen("/Users/perpeternal/Documents/Sketch/data/in.dat", "r", stdin);
    // freopen("/Users/perpeternal/Documents/Sketch/data/out.dat", "w", stdout);
#endif
    //ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int n,m,a[N];
    while(~in(n)){
        in(m);
        for0(i,n)in(a[i]);
        sort(a,a+n,cmp);
        int remain=m;
        int i=1;
        for(;i<n;i++){
            if(i*(a[i-1]-a[i])<=remain){
                remain-=i*(a[i-1]-a[i]);
            }else break;
        }
        fraction ans=fraction(i*a[i-1]*a[i-1]-2*remain*a[i-1],1)+fraction(remain*remain,i);
        for(;i<n;i++){
            ans=ans+fraction(a[i]*a[i],1);
        }
        ans=ans/fraction(m*m,1);
        ans.display();
    }
    return 0;
}