2019 Multi-University Training Contest 2

Posted 2022-05-09 mountaink

 

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1005  Everything Is Generated In Equal Probability

 

//#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<string>
#include<math.h>
#include<cmath>
#include<time.h>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<algorithm>
#include<numeric>
#include<stack>
#include<bitset>
#include<unordered_map>
const int maxn = 0x3f3f3f3f;
const double EI = 2.71828182845904523536028747135266249775724709369995957496696762772407663035354594571382178525166427;
const double PI = 3.141592653589793238462643383279;
//#ifdef TRUETRUE
//#define gets gets_s
//#endif
using namespace std;
long long p = 998244353;
long long quick(long long a, int b, int c)

    long long ans = 1;
    a = a % c;
    while (b != 0)
    
        if (b & 1)
        
            ans = (ans * a) % c;
        
        b >>= 1;
        a = (a * a) % c;
    
    return ans;

struct s

    long long a, b,ans;
z[3010];
long long ans[3010], aa[3010];
int main(void)

    int i, j, k;
    long long zz = 2;
    z[1].a = 0;
    z[1].ans = 0;
    long long z3 = quick(3, p - 2, p);
    //printf("  %lld\n",z3);
    for (i = 2; i <= 3000; i++)
    
        z[i].a = z[i - 1].a + zz;
        z[i].a %= p;

        z[i].ans = (z[i].a * z3) % p;

        zz += 2;
        zz %= p;
    
    ans[1] = 0;
    aa[1] = 0;
    for (i = 2; i <= 3000; i++)
    
        ans[i] = (ans[i - 1] + z[i].ans) % p;
        aa[i] = (ans[i] * quick(i, p - 2, p)) % p;
    
    int N;
    while (~scanf("%d", &N))
    
        printf("%lld\n", aa[N]);
    
    return 0;

View Code

 

1010 Just Skip The Problem

 

//#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<string>
#include<math.h>
#include<cmath>
#include<time.h>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<algorithm>
#include<numeric>
#include<stack>
#include<bitset>
#include<unordered_map>
const int maxn = 0x3f3f3f3f;
const double EI = 2.71828182845904523536028747135266249775724709369995957496696762772407663035354594571382178525166427;
const double PI = 3.141592653589793238462643383279;
//#ifdef TRUETRUE
//#define gets gets_s
//#endif
using namespace std;
long long p = 1e6 + 3;
int main(void)

    long long n,i,ans;
    while (~scanf("%lld",&n))
    
        if (n >= p)
        
            printf("0\n");
            continue;
        
        ans = 1;
        for (i = 1;i <= n;i++)
        
            ans *= i;
            ans %= p;
        
        printf("%lld\n",ans);
    
    return 0;

View Code

 

1011 Keen On Everything But Triangle

题意：给出n根火柴，q次询问，每次询问[l,r]区间内的火柴组成三角形的最长周长是多少。

思路：考虑暴力，必定是将l到r区间内的所有数字排序，然后从大到小依次check相邻的三根火柴能否组成三角形。

　　考虑优化，由斐波那契数列的性质可得，45根1e9范围内的火柴必定能组成一个三角形（最坏情况是1,1,2,3,5，第45项就超过1e9了），所以线段树区间维护46个最大值，然后做上面的check就可以了。

#include<bits/stdc++.h>
#define clr(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
const int maxn=200010;
struct node
    int l,r,size;
    ll a[50];
tr[maxn<<2];
ll val[maxn];
int n,q;
inline ll rd()

    ll x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘)if(ch==‘-‘)f=-1;ch=getchar();
    while(ch>=‘0‘&&ch<=‘9‘)x=x*10+ch-‘0‘;ch=getchar();
    return x*f;

inline void pushup(int o,int oa,int ob)
    int i=1,j=1,k=0;
    while(++k<=tr[o].size)
        if(i<=tr[oa].size&&tr[oa].a[i]>=tr[ob].a[j])
            tr[o].a[k]=tr[oa].a[i];
            i++;
        else
            tr[o].a[k]=tr[ob].a[j];
            j++;
        
    

inline void build(int o,int l,int r)
    tr[o].size=min(r-l+1,46);
    for(int i=1;i<=tr[o].size;i++)
        tr[o].a[i]=0;
    
    if(l==r)
        tr[o].a[1]=val[l];
        return;
    
    int mid=(l+r)>>1;
    build(o<<1,l,mid);
    build(o<<1|1,mid+1,r);
    pushup(o,o<<1,o<<1|1);

ll ans[50],temp[50];
int top=0;
inline void query(int o,int l,int r,int ql,int qr)
    if(ql<=l&&r<=qr)
        int si=min(46,top+tr[o].size);
        int i=1,j=1,k=0;
        while(++k<=si)
            if(i<=top&&ans[i]>=tr[o].a[j])
                temp[k]=ans[i];
                i++;
            else
                temp[k]=tr[o].a[j];
                j++;
            
        
        top=si;
        for(int i=1;i<=top;i++)
            ans[i]=temp[i];
        
        for(int i=top+1;i<=46;i++)
            ans[i]=0;
        
        return;
    
    int mid=(l+r)>>1;
    if(ql<=mid)query(o<<1,l,mid,ql,qr);
    if(mid<qr)query(o<<1|1,mid+1,r,ql,qr);

int main()
    while(cin>>n>>q)
        for(int i=1;i<=n;i++)
//            scanf("%lld",&val[i]);
            val[i]=rd();
        
        build(1,1,n);
        while(q--)
            int l,r;
            top=0;
            l=rd(),r=rd();
            query(1,1,n,l,r);
            ll an=-1;
            for(int i=1;i<=top-2;i++)
                if(ans[i]<ans[i+1]+ans[i+2])
                    an=ans[i]+ans[i+1]+ans[i+2];
                    break;
                
            
            printf("%lld\n",an);
        
    

View Code

 

1012 Longest Subarray

题意：给出n个数字，求一个最长子串，要求这个串包含的数字的次数必定大于等于k。

思路：先将整体的字符串，次数小于k的数全部去掉，得到若干个不连续的子串，对这些子串递归做这个工作。调一下参数，递归到30层时直接return。

（这个做法可以被hack，但数据比较难造，赌出题人没有卡这个做法，队友tql）

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dep(i,a,b) for(int i=b;i>=a;i--)
using namespace std;
#define ll long long
const int N=3e5+5;
const int mod = 998244353;
int a[301010],c[301010],q[301010];
int n,C,k,ans;
ll rd()

    ll x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘)if(ch==‘-‘)f=-1;ch=getchar();
    while(ch>=‘0‘&&ch<=‘9‘)x=x*10+ch-‘0‘;ch=getchar();
    return x*f;

void solve(int l,int r,int dep)

    if(dep>=30) return;
    //printf("l=%d r=%d\n",l,r);
    if(r-l+1<=ans) return;
    int flag=0,pre,cnt=0;
    rep(i,l,r) c[a[i]]=0;
    rep(i,l,r) ++c[a[i]];
    q[++cnt]=l-1;
    rep(i,l,r) 
    
        if(c[a[i]]<k) 
        
            q[++cnt]=i;
            flag=1;
        
    
    q[++cnt]=r+1;
    if(flag==0)
    
        ans=r-l+1;    
        
    rep(i,1,cnt) solve(q[i]+1,q[i+1]-1,dep+1);

int main()

//    freopen("1.in","r",stdin);
//    freopen("1.out","w",stdout);
    while(~scanf("%d%d%d",&n,&c,&k))
        
        ans=0;
        rep(i,1,n) a[i]=rd();
        solve(1,n,1);
        printf("%d\n",ans);
    

View Code