題意：有n個禮物，1~n。m個區間，第i個為li~ri。每次隨機三個區間，取走 no less than max(li,lj,lk) and no larger than間的禮物，問取走禮物個數的期望

解法一：線段樹區間更新，對大於三的點C（n，3），後求和。值為n說明有n官網區間可以覆蓋這個點，從中選三個。

程式碼：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
#define mem(a,b) memset((a),(b),sizeof((a)))
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
#define root 1 , n , 1
#define ll long long

void RI (int& x){
    x = 0;
    char c = getchar ();
    while (c == ' '||c == '\n')    c = getchar ();
    bool flag = 1;
    if (c == '-'){
        flag = 0;
        c = getchar ();
    }
    while (c >= '0' && c <= '9'){
        x = x * 10 + c - '0';
        c = getchar ();
    }
    if (!flag)    x = -x;
}
void RII (int& x, int& y){RI (x), RI (y);}
void RIII (int& x, int& y, int& z){RI (x), RI (y), RI (z);}

const int maxn = 50010;
ll add[maxn<<2];
ll sum[maxn<<2];
void PushUp(int rt) {
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void PushDown(int rt,int m) {
    if (add[rt]) {
        add[rt<<1] += add[rt];
        add[rt<<1|1] += add[rt];
        sum[rt<<1] += add[rt] * (m - (m >> 1));
        sum[rt<<1|1] += add[rt] * (m >> 1);
        add[rt] = 0;
    }
}
void update(int L,int R,int c,int l,int r,int rt) {
    if (L <= l && r <= R) {
        add[rt] += c;
        sum[rt] += (ll)c * (r - l + 1);
        return ;
    }
    PushDown(rt , r - l + 1);
    int m = (l + r) >> 1;
    if (L <= m) update(L , R , c , lson);
    if (m < R) update(L , R , c , rson);
    PushUp(rt);
}
ll query(int L,int R,int l,int r,int rt) {
    if (L <= l && r <= R) {
        return sum[rt];
    }
    PushDown(rt , r - l + 1);
    int m = (l + r) >> 1;
    ll ret = 0;
    if (L <= m) ret += query(L , R , lson);
    if (m < R) ret += query(L , R , rson);
    return ret;
}
int mon[maxn];
ll gcd(ll a, ll b){
    return b==0?a:gcd(b,a%b);
}
int main() {
    //freopen("test.txt","r",stdin);
    int T;
    RI(T);
    while(T --){
        int n,m;
        RII(n,m);
        mem(sum,0);mem(add,0);
        for(int i = 1;i <= n;i ++){
            RI(mon[i]);
        }
        for(int i = 0;i < m;i ++){
            int l,r;
            RII(l,r);
            update(l,r,1,1,n,1);
        }
        ll ans = 0;
        for(int i = 1;i <= n;i ++){
            ll nn = query(i,i,1,n,1);//cout<<nn<<' ';
            if(nn >= 3)ans += (ll)mon[i]*nn*(nn-1)*(nn-2)/6;
        }

        if(m <= 2){cout<<0<<endl;continue;}
        ll mu = (ll)m*(m-1)*(m-2)/6;
        if(ans == 0){cout<<0<<endl;continue;}

        ll gg = gcd(ans,mu);
        ans /= gg;mu /= gg;
        if(mu == 1){
            cout<<ans<<endl;
        }
        else printf("%I64d/%I64d\n",ans,mu);
    }
    return 0;
}
 

不用線段樹，對於每個區間，起點加一，終點減一。具體見程式碼。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
#define mem(a,b) memset((a),(b),sizeof((a)))
#define ll long long

void RI (int& x){
    x = 0;
    char c = getchar ();
    while (c == ' '||c == '\n')    c = getchar ();
    bool flag = 1;
    if (c == '-'){
        flag = 0;
        c = getchar ();
    }
    while (c >= '0' && c <= '9'){
        x = x * 10 + c - '0';
        c = getchar ();
    }
    if (!flag)    x = -x;
}
void RII (int& x, int& y){RI (x), RI (y);}
void RIII (int& x, int& y, int& z){RI (x), RI (y), RI (z);}


const int maxn = 50010;
int mon[maxn];
int num[maxn];
int main() {
    //freopen("test.txt","r",stdin);
    int T;
    RI(T);
    while(T --){
        int n,m;
        RII(n,m);
        mem(num,0);
        for(int i = 1;i <= n;i ++){
            RI(mon[i]);
        }
        for(int i = 0;i < m;i ++){
            int l,r;
            RII(l,r);
            num[l] ++;num[r+1] --;
        }
        ll ans = 0;
        int tmp = 0;
        for(int i = 1;i <= n;i ++){
            tmp += num[i];//cout<<tmp<<' ';
            if(tmp >= 3)ans += (ll)mon[i]*tmp*(tmp-1)*(tmp-2)/6;
        }

        if(m <= 2){cout<<0<<endl;continue;}
        ll mu = (ll)m*(m-1)*(m-2)/6;
        if(ans == 0){cout<<0<<endl;continue;}

        ll gg = __gcd(ans,mu);
        ans /= gg;mu /= gg;
        if(mu == 1){
            cout<<ans<<endl;
        }
        else printf("%I64d/%I64d\n",ans,mu);
    }
    return 0;
}