2019hdu第二场

10：签到求n!取模

#include <iostream>
#include <iterator>
#include <algorithm>
typedef long long ll;
using namespace std;
const ll mod=1e6+;
ll n;
ll res[mod+];
int main(){
    res[]=%mod;
    res[]=%mod;
    for(int i=;i<=mod+;i++){
        res[i]=(res[i-]*i)%mod;
    }
    while(scanf("%lld",&n)!=EOF){
        if(n>=mod) printf("0\n");
        else printf("%lld\n",res[n]);
    }
    return ;
}

11：签到，主席树维护a[]数组，然后对于每个q，去询问区间第1大，第2大，第3大，如果不能组成三角形，则询问区间第4大，依次类推，询问次数最多为O(logn)，总复杂度O(qlogn*logn)

/*
主席树维护区间[l,r]
查找第1大，第2大，第3大，第4大
*/
#include<bits/stdc++.h>
using namespace std;
#define maxn 100005
#define ll long long
ll a[maxn],n,b[maxn];

int tot,m,rt[maxn];
struct Node{int lc,rc,sum;}t[maxn*];
int update(int last,int l,int r,int pos){
    int now=++tot;
    t[now]=t[last];t[now].sum++;
    if(l==r)return now;
    int mid=l+r>>;
    if(pos<=mid)
        t[now].lc=update(t[last].lc,l,mid,pos);
    else t[now].rc=update(t[last].rc,mid+,r,pos);
    return now;
}
int query(int st,int ed,int l,int r,int k){
    if(l==r)return l;
    int mid=l+r>>;
    int sum=t[t[ed].lc].sum-t[t[st].lc].sum;
    if(k<=sum)return query(t[st].lc,t[ed].lc,l,mid,k);
    else return query(t[st].rc,t[ed].rc,mid+,r,k-sum);
}
int build(int l,int r){
    int now=++tot;
    t[now].lc=t[now].rc=t[now].sum=;
    if(l==r)return now;
    int mid=l+r>>;
    t[now].lc=build(l,mid);
    t[now].rc=build(mid+,r);
    return now;
}

int main(){
    int q;
    while(scanf("%d%d",&n,&q)==){
        for(int i=;i<=n;i++){
            scanf("%d",&a[i]);
            b[i]=a[i];
        }
        sort(b+,b++n);
        m=unique(b+,b++n)-(b+);
//cout<<m<<endl;
        for(int i=;i<=n;i++)
            a[i]=lower_bound(b+,b++m,a[i])-b;

        tot=;
        rt[]=build(,m);
        for(int i=;i<=n;i++)
            rt[i]=update(rt[i-],,m,a[i]);
        int l,r;
        while(q--){
            scanf("%d%d",&l,&r);
            if(r-l+<){puts("-1");continue;}
            int len=r-l+;
            ll x=b[query(rt[l-],rt[r],,m,len)],y=b[query(rt[l-],rt[r],,m,len-)],z=b[query(rt[l-],rt[r],,m,len-)];
            len-=;
            while(x>=y+z){
                if(len==)break;
                x=y,y=z;z=b[query(rt[l-],rt[r],,m,len)];
                len--;
            }
            if(x>=y+z)puts("-1");
            else cout<<x+y+z<<'\n';
        }

    }
} 

12：队友过的规律题，从[1,n]里随机选一个数丢进函数里，问最后答案的期望值

实质是一个概率dp，可以通过n^2的转移来做，当然也可以直接找规律推出线性通项公式做

这是比较好的题解（比官方题解容易懂）https://www.cnblogs.com/isakovsky/p/11247444.html

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <stack>
const int MOD = ;
#define rep(i,n,m) for(int i=n;i<=m;++i)
const double eps = 1e-;
#define ll long long
using namespace std;
const int maxn = ;
const int maxm = 2e5 + ;
const int inf =  << ;
typedef pair<int, int> P;
#define zero(a) fabs(a)<eps
inline int read()
{
    int x = , f = ;
    char ch = getchar();
    while (ch < '' || ch > '')
    {
        if (ch == '-')
            f = -;
        ch = getchar();
    }
    while (ch >= '' && ch <= '')
    {
        x =  * x + ch - '';
        ch = getchar();
    }
    return x * f;
}

ll quick_mod(ll x, ll n) {
    ll res = ;
    while (n) {
        if (n & ) res = res * x % MOD;
        x = x * x % MOD;
        n = n >> ;
    }
    return res;
}
ll a[];
void init()
{
    a[] = a[] = ;
    a[] = ;
    for (int i = ; i <= ; ++i)
        a[i] = a[i - ] + (i - ) * ;
    for (int i = ; i <= ; ++i)
        a[i] += a[i - ];
}
int main()
{
    ll n;
    init();
    while (~scanf("%lld", &n))
    {
        if (n == ) { cout << "" << endl; continue; }
        ll a1 = a[n];
        ll b =  * n;
        ll x = quick_mod(b, MOD - );
        cout << a1 % MOD * x % MOD<<endl;
    }
}