2019 Multi-University Training Contest 7

Final Exam

题意：一共有n道题，需要做对k道题，每个题有一个分数，要想做对这道题，需要准备的时间是这道题的分数加1，给出n道题的总分数m，问你做对k道题花费的最少时间，还有就是，你学了哪儿道题，出题人就会想办法阻止你，比如你学一道题1分钟，他会把这道题目的分数提高到>=1分，但是满足最后的总分数是m。

 

题解：他阻止我分数最小的n-k+1个，然后把剩下的k-1个变成前面n-k个里面“最大的数(相等或者加一)”（正好整除，就是前面的数加1，不整除就是前面的向下取整加一，然后把多于的平均分到前面的上面就好），然后前面的数再加1（正好整除随便分到前面任意一个数，不整除可以加到前面的没有平均分到的那个位置），这样保证前面n-k+1个的总和>m，这样总时间会更小.

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
ll n,m,k,k2,k1;
int main()
{
    int _;
    scanf("%d",&_);
    while (_--)
    {
        scanf("%lld%lld%lld",&n,&m,&k);
        printf("%lld
",m+1+(m/(n-k+1)+1)*(k-1));
    }
}

Kejin Player

E[i]+a[i]=E[i+1]*p[i]+E[x[i]]*(1-p[i])

#include <bits/stdc++.h>
 
using namespace std;
const int mod=1e9+7;
typedef long long ll;
const int maxn=500010;
ll n,q,r[maxn],s[maxn],x[maxn],a[maxn],p[maxn],f[maxn];
 
ll pow_mod(ll a,ll b)
{
    ll res=1;
    while (b)
    {
        if (b&1)
        {
            res=res*a%mod;
        }
        a=a*a%mod;
        b>>=1;
    }
    return res;
}
 
ll inv(ll x)
{
    return pow_mod(x,mod-2);
}
inline ll read()
{
    ll res=0,f=1;
    char ch=getchar();
    while (!isdigit(ch))
    {
        if (ch=='-')
        {
            f=-f;
        }
        ch=getchar();
    }
    while (isdigit(ch))
    {
        res=(res<<3)+(res<<1)+ch-'0';
        ch=getchar();
    }
    return f*res;
}
int main()
{
    int _;
    scanf("%d",&_);
    while (_--)
    {
        scanf("%lld%lld",&n,&q);
        for (int i=1; i<=n; i++)
        {
            r[i]=read();s[i]=read();x[i]=read();a[i]=read();
            p[i]=r[i]*inv(s[i])%mod;
        }
        for (int i=2; i<=n+1; i++)
        {
            f[i]=(f[i-1]+a[i-1]%mod+f[x[i-1]]*(mod-(1-p[i-1]+mod))%mod)*inv(p[i-1])%mod;
        }
 
        while (q--)
        {
            ll l,r;
            scanf("%lld%lld",&l,&r);
            printf("%lld
",(f[r]-f[l]+mod)%mod);
        }
    }
}

A + B = C

 

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int maxn = 5e5 + 10,limit = 3e5 + 6;
int dis[maxn];

int solve(int c[],int lenc,int a[],int lena,int b[],int lenb)
{
    for (int i=0; i<limit; i++)
    {
        if (c[i]!=a[i])
        {
            if (c[i]<a[i])
            {
                return -1;
            }
            break;
        }
    }
    for (int i=0; i<limit; i++)
    {
        dis[i]=c[i]-a[i];
    }
    for (int i=limit-1; i; i--)
    {
        if (dis[i]<0)
        {
            dis[i-1]--;
            dis[i]+=10;
        }
    }
    int cur=0;
    while (dis[cur]==0&&cur<limit)
    {
        cur++;
    }
    if (cur==limit)
    {
        return -1;
    }
    for (int i=0; i<lenb; i++)
    {
        if (dis[cur+i]!=b[i])
        {
            return -1;
        }
    }
    for (int i=lenb; i<limit; i++)
    {
        if (dis[i+cur])
        {
            return -1;
        }
    }
    return limit-cur-lenb;
}
int a[maxn], b[maxn], c[maxn];
char s[maxn];

int main()
{
    int lena,lenb,lenc,_;
    scanf("%d",&_);
    while (_--)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(c,0,sizeof(c));
        scanf("%s",s);
        for (int i=0; s[i]; i++,lena=i)
        {
            a[i+1]=s[i]-'0';
        }
        scanf("%s",s);
        for (int i=0; s[i]; i++,lenb=i)
        {
            b[i+1]=s[i]-'0';
        }
        scanf("%s",s);
        for (int i=0; s[i]; i++,lenc=i)
        {
            c[i+1]=s[i]-'0';
        }
        int res;
        if ((res=solve(c+1,lenc,a+1,lena,b+1,lenb))!=-1)
        {
            printf("%d %d %d
",limit-lena,res,limit-lenc);
            continue;
        }
        if ((res=solve(c+1,lenc,a,lena+1,b+1,lenb))!=-1)
        {
            printf("%d %d %d
",limit-lena-1,res,limit-lenc);
            continue;
        }
        if ((res=solve(c+1,lenc,b+1,lenb,a+1,lena))!=-1)
        {
            printf("%d %d %d
",res,limit-lenb,limit-lenc);
            continue;
        }
        if ((res=solve(c+1,lenc,b,lenb+1,a+1,lena))!=-1)
        {
            printf("%d %d %d
",res,limit-lenb-1,limit-lenc);
            continue;
        }
        printf("-1
");
    }
    return 0;
}

Just Repeat

 

 

#pragma GCC optimize(2)
#include <bits/stdc++.h>
 
using namespace std;
 
unsigned long long k1, k2, mod;
int n,m;
const int maxn=1e5+5;
int a[maxn],b[maxn],p,cnt,suma,sumb;
 
unsigned long long rng()
{
    unsigned long long k3 = k1, k4 = k2;
    k1 = k4;
    k3 ^= k3 << 23;
    k2 = k3 ^ k4 ^ (k3 >> 17) ^ (k4 >> 26);
    return k2 + k4;
}
struct node
{
    int num,a,b;
    bool operator<(const node &b)const
    {
        return num>b.num;
    }
} c[maxn];
 
inline int read()
{
    int res=0,f=1;
    char ch=getchar();
    while (!isdigit(ch))
    {
        if (ch=='-')
        {
            f=-f;
        }
        ch=getchar();
    }
    while (isdigit(ch))
    {
        res=(res<<3)+(res<<1)+ch-'0';
        ch=getchar();
    }
    return f*res;
}
 
int main()
{
    int _;
    scanf("%d",&_);
    while (_--)
    {
        unordered_map <int,int> m1,m2;
        suma=sumb=cnt=0;
        scanf("%d%d%d",&n,&m,&p);
        if (p==1)
        {
            for (register int i=0; i<n; ++i)
            {
                a[i]=read();
            }
            for (register int i=0; i<m; ++i)
            {
                b[i]=read();
            }
        }
        else
        {
            scanf("%llu%llu%llu",&k1,&k2,&mod);
            for (int i = 0; i < n; ++i)
                a[i] = rng() % mod;
            scanf("%llu%llu%llu",&k1,&k2,&mod);
            for (int i = 0; i < m; ++i)
                b[i] = rng() % mod;
        }
        for (int i=0; i<n; i++)
        {
            m1[a[i]]++;
        }
        for (int i=0; i<m; i++)
        {
            m2[b[i]]++;
        }
        for (register int i=0; i<m; ++i)
        {
            if (m1[b[i]]>0)
            {
                ++cnt;
                c[cnt].num=m1[b[i]]+m2[b[i]];
                c[cnt].a=m1[b[i]];
                c[cnt].b=m2[b[i]];
                m1[b[i]]=-1;
            }
            else if (m1[b[i]]!=-1&&m2[b[i]]!=-1)
            {
                sumb+=m2[b[i]];
                m2[b[i]]=-1;
            }
        }
         
        for (register int i=0; i<n; i++)
        {
            if (m2[a[i]]==0)
            {
                suma+=m1[a[i]];
                m2[a[i]]=-1;
            }
        }
        sort(c+1,c+cnt+1);
        for (register int i=1; i<=cnt; ++i)
        {
            if (i&1)
            {
                suma+=c[i].a;
            }
            else
            {
                sumb+=c[i].b;
            }
        }
 
        if (suma>sumb)
        {
            printf("Cuber QQ
");
        }
        else
        {
            printf("Quber CC
");
        }
    }
    return 0;
}
 

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