题解：2018 Multi-University Training Contest 1

垫底进队之后第一次集训。

Maximum Multiple

3的倍数拆1,1,1，4的倍数拆1,1,2，优先拆3的倍数。

#include<stdio.h>
long long t,n;
int main()
{
    for(scanf("%lld",&t); t--;
            printf("%lld\n",n%3==0?n*n*n/27:
                   n%4==0?n*n*n/32:-1))
        scanf("%lld",&n);
}

Balanced Sequence

具体思想是把每个串按删去可以自匹配的括号后的剩余串（形如)))((）排序：
左括号多于右括号的串排在左括号少于右括号的串前面；
同类型的串中，左括号多于右括号的串右括号少的放前面，右括号多于左括号的串左括号少的放前面。
现场谜之排序WA了十一发…

#include<bits/stdc++.h>
using namespace std;
const int N=100009;
struct Node
{
    int l,r;
    bool operator<(const Node &t)const
    {
        return r<l?(t.r<t.l?r<t.r:1):(t.r<t.l?0:l>t.l);
    }
} v[N];
char s[N];
int t,n,ans;
int main()
{
    for(scanf("%d",&t); t--;)
    {
        scanf("%d",&n);
        for(int i=ans=0; i<n; ++i)
        {
            scanf("%s",s);
            for(int j=v[i].l=v[i].r=0; s[j]; ++j)
            {
                if(s[j]=='(')++v[i].l;
                else if(v[i].l)--v[i].l,++ans;
                else ++v[i].r;
            }
        }
        sort(v,v+n);
        for(int i=0,cnt=0; i<n; ++i)
        {
            ans+=min(cnt,v[i].r);
            cnt+=v[i].l-min(cnt,v[i].r);
        }
        printf("%d\n",ans*2);
    }
}

Triangle Partition

把所有点按照横坐标-纵坐标的字典序排序，选取依次相邻三项。

#include<bits/stdc++.h>
using namespace std;
struct Coord:pair<int,int>
{
    int id;
} p[10009];
int t,n;
int main()
{
    for(scanf("%d",&t); t--;)
    {
        scanf("%d",&n);
        for(int i=0; i<3*n; ++i)
            scanf("%d%d",&p[i].first,&p[i].second),p[i].id=i;
        sort(p,p+3*n);
        for (int i=0; i<3*n; ++i)
            printf("%d%c",p[i].id+1,i%3!=2?' ':'\n');
    }
}

Distinct Values

队友写的。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;

const int maxn=100000+10;

pair<int,int> a[maxn];
int ans[maxn];
int t,n,m;
priority_queue<int,vector<int>,greater<int> > q;

int main()
{
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d%d",&n,&m);
        memset(ans,0,sizeof(ans));
        if (m==0)
        {
            for (int i=1;i<n;i++) printf("1 ");
            printf("1\n");
            continue;
        }
        for (int i=1;i<=m;i++)
         scanf("%d%d",&a[i].first,&a[i].second);
        sort(a+1,a+m+1);
        while (!q.empty()) q.pop();
    //  for (int i=1;i<=m;i++) printf("!%d %d\n",a[i].first,a[i].second);
        for (int i=1;i<a[1].first;i++) ans[i]=1;
        for (int i=a[1].first;i<=a[1].second;i++) ans[i]=i-a[1].first+1;

        int l=a[1].first,r=a[1].second,mx=ans[a[1].second];
        for (int i=2;i<=m;i++)
        {
            if (a[i].second<=r) continue;
            if (a[i].first<=r)
            {
                for (int j=l;j<a[i].first;j++) q.push(ans[j]);
                for (int j=r+1;j<=a[i].second;j++)
                 if (!q.empty()) ans[j]=q.top(),q.pop();
                  else ans[j]=++mx;
            }
            else
            {
                while (!q.empty()) q.pop();
                mx=0;
                for (int j=a[i].first;j<=a[i].second;j++) ans[j]=++mx;
            }
            l=a[i].first; r=a[i].second;
        }
        for (int i=1;i<=n;i++)
         if (ans[i]==0) ans[i]=1;
        for (int i=1;i<n;i++) printf("%d ",ans[i]);
        printf("%d\n",ans[n]);
    }
}

Maximum Weighted Matching

//假装会写

Period Sequence

//假装会写

Chiaki Sequence Revisited

队友$O(\log\log N)$的二分强行跑掉！太强啦！

#include <cstdio>

typedef long long ll;
const ll mod=1e9+7;

int t;
ll n,v,ans,u;

inline ll geti(ll x)
{
    ll l=x/2-1,r=x;
    while (r-l>1)
    {
        ll mid=(l+r)/2,k=mid;
        ll g=0;
        while (k>0)
        {
            g+=k;
            k/=2;
        }
    //  printf("%lld %lld %lld %lld\n",l,r,mid,g);
        if (g<x) l=mid;
         else r=mid;
    }
    return r;
}

inline ll cal(ll x)
{
    ll ret=0,t=1;
    u=0;
    while (x>0) 
    {
        if (x&1) ret=(ret+((x%mod)*((x+1)/2%mod)%mod)*t)%mod;
         else ret=(ret+(((x+1)%mod)*((x/2)%mod)%mod)*t)%mod;
        //printf("ret=%lld\n",ret);
        u+=x;
        t=(t*2)%mod;
        x/=2;
    }
    return ret;
}

int main()
{
    scanf("%d",&t);
    while (t--)
    {
        scanf("%lld",&n);
        v=geti(n-1);
    //  printf("%lld ",v);
        ans=cal(v-1);
        ans=(ans+v*(n-1-u)+1)%mod;
        printf("%lld\n",ans);
    }
    return 0;
}

RMQ Similar Sequence

学习一个一队大爷的分治做法。
dfs写起来奥妙重重…递归要尽量写在函数尾，否则爆栈。

#include<bits/stdc++.h>
#define log2(n) LOG[n]
#define mul(a,b,m) ((long long)(a)*(b)%(m))
using namespace std;
const int N=1e6+7,M=1e9+7;
pair<int,int> st[20][N];
int t,n,ans,LOG[N]= {-1},FAC[N]= {1},IFAC[N];
void dfs(int l,int r)
{
    if(l>r)return;
    int m=log2(r-l+1);
    m=min(st[m][l],st[m][r+1-(1<<m)]).second;
    if(l<m&&m<r)ans=mul(mul(mul(ans,FAC[r-l],M),IFAC[m-l],M),IFAC[r-m],M);
    dfs(l,m-1),dfs(m+1,r);
}
int main()
{
    for(int i=IFAC[N-1]=1; i<N; ++i)LOG[i]=LOG[i>>1]+1,FAC[i]=mul(FAC[i-1],i,M),st[0][i].second=i;
    for(int a=FAC[N-1],b=M-2; b; b>>=1,a=mul(a,a,M))if(b&1)IFAC[N-1]=mul(IFAC[N-1],a,M);
    for(int i=N-1; i; --i)IFAC[i-1]=mul(IFAC[i],i,M);
    for(scanf("%d",&t); t--; printf("%d\n",ans))
    {
        scanf("%d",&n);
        for(int i=0; i<n; ++i)scanf("%d",&st[0][i].first),st[0][i].first*=-1;
        for(int k=0; k<log2(n); ++k)for(int i=0; i+(1<<k)<n; ++i)st[k+1][i]=min(st[k][i],st[k][i+(1<<k)]);
        ans=mul(mul(n,IFAC[2],M),IFAC[n],M),dfs(0,n-1);
    }
}

Lyndon Substring

//假装会写

Turn Off The Light

//假装会写

Time Zone

读到的double乘十后要四舍五入，且istringstream速度太慢过不了。

#include<stdio.h>
char s[9];
double d;
int t,a,b;
int main()
{
    for(scanf("%d",&t); t--; printf("%02d:%02d\n",(a+b/60)%24,b%60))
    {
        scanf("%d%d%s",&a,&b,&s);
        sscanf(s+3,"%lf",&d);
        if(d-=8,d<0)d+=24;
        b+=6*(int)(d*10+0.5);
    }
}