2018 ACM 网络选拔赛 南京赛区

Posted 2021-01-07 cmyg

 

A. An Olympian Math Problem

 1 #include <cstdio>
 2 #include <cstdlib>
 3 #include <cmath>
 4 #include <cstring>
 5 #include <time.h>
 6 #include <string>
 7 #include <set>
 8 #include <map>
 9 #include <list>
10 #include <stack>
11 #include <queue>
12 #include <vector>
13 #include <bitset>
14 #include <ext/rope>
15 #include <algorithm>
16 #include <iostream>
17 using namespace std;
18 #define ll long long
19 #define minv 1e-6
20 #define inf 1e9
21 #define pi 3.1415926536
22 #define nl 2.7182818284
23 const ll mod=1e9+7;//998244353
24 const int maxn=1e5+10;
25 
26 
27 int main()
28 {
29     int t;
30     ll n;
31     scanf("%d",&t);
32     while (t--)
33     {
34         scanf("%lld",&n);
35         printf("%lld
",n-1);
36     }
37     return 0;
38 }

 

C. GDY

  1 #include <cstdio>
  2 #include <cstdlib>
  3 #include <cmath>
  4 #include <cstring>
  5 #include <time.h>
  6 #include <string>
  7 #include <set>
  8 #include <map>
  9 #include <list>
 10 #include <stack>
 11 #include <queue>
 12 #include <vector>
 13 #include <bitset>
 14 #include <ext/rope>
 15 #include <algorithm>
 16 #include <iostream>
 17 using namespace std;
 18 #define ll long long
 19 #define minv 1e-6
 20 #define inf 1e9
 21 #define pi 3.1415926536
 22 #define nl 2.7182818284
 23 const ll mod=1e9+7;//998244353
 24 const int maxn=2e2+10;
 25 const int maxm=2e4;
 26 
 27 int s[maxm],a[maxn][maxm],g[maxn];
 28 int value[14]={0,12,13,1,2,3,4,5,6,7,8,9,10,11};
 29 
 30 int cmp(int x,int y)
 31 {
 32     return value[x]>value[y];
 33 }
 34 
 35 int main()
 36 {
 37     int t,T,n,m,num,i,j,sum;
 38     int x,y,z;
 39     scanf("%d",&t);
 40     for (T=1;T<=t;T++)
 41     {
 42         memset(g,0,sizeof(g));
 43 
 44         scanf("%d%d",&n,&m);
 45         for (i=0;i<m;i++)
 46             scanf("%d",&s[i]);
 47         num=0;
 48         for (i=0;i<n;i++)
 49         {
 50             for (j=0;j<5;j++)
 51             {
 52                 if (num==m)
 53                     break;
 54                 a[i][j]=s[num++];
 55             }
 56             sort(a[i],a[i]+j,cmp);
 57             g[i]=j;
 58         }
 59 
 60         z=a[0][g[0]-1];//previous number
 61         g[0]--;
 62         y=1;//has y persons
 63         x=1;//pos
 64         while (1)
 65         {
 66             for (i=g[x]-1;i>=0;i--)
 67                 if (value[z]+1==value[a[x][i]] || (a[x][i]==2 && z!=2))
 68                     break;
 69 
 70             if (i!=-1)
 71             {
 72                 z=a[x][i];
 73                 a[x][i]=0;
 74                 sort(a[x],a[x]+g[x],cmp);
 75                 g[x]--;
 76                 if (g[x]==0)
 77                     break;
 78                 y=1;
 79             }
 80             else if (y!=n-1)
 81                 y++;
 82             else
 83             {
 84                 x=(x+1)%n;
 85                 for (i=x;;i=(i+1)%n)
 86                 {
 87                     if (num==m)
 88                         break;
 89                     else
 90                     {
 91                         a[i][g[i]++]=s[num++];
 92                         sort(a[i],a[i]+g[i],cmp);
 93                     }
 94                     if (i==(x-1+n)%n)
 95                         break;
 96                 }
 97 
 98                 z=a[x][g[x]-1];
 99                 g[x]--;
100                 if (g[x]==0)
101                     break;
102                 y=1;
103             }
104 
105             x=(x+1)%n;
106         }
107         printf("Case #%d:
",T);
108         for (i=0;i<n;i++)
109             if (i==x)
110                 printf("Winner
");
111             else
112             {
113                 sum=0;
114                 for (j=0;j<g[i];j++)
115                     sum+=a[i][j];
116                 printf("%d
",sum);
117             }
118     }
119     return 0;
120 }
121 /*
122 10
123 2 6
124 3 5 7 9 11 4
125 
126 3 20
127 2 2 2 2 2 2 2 2 2 2
128 2 2 2 2 2 2 2 2 2 2
129 
130 3 19
131 2 2 2 2 2 2 2 2 2 2
132 2 2 2 2 2 2 2 2 2
133 
134 3 20
135 3 2 2 2 2 2 2 2 2 2 2
136 2 2 2 2 2 2 2 2 2
137 
138 2 10
139 3 4 5 6 7 2 3 4 5 6
140 
141 3 15
142 3 4 5 6 7 12 12 12 12 12 13 13 13 13 13
143 
144 3 11
145 1 2 3 4 5 6 7 8 9 10 11
146 
147 3 13
148 1 2 3 4 5 6 7 8 9 10 11 12 13
149 
150 1
151 3 17
152 1 2 3 4 5 6 7 8 9 10 11 12 13 12 11 10 9
153 
154 */

 

E. AC Challenge

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <time.h>
#include <string>
#include <set>
#include <map>
#include <list>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <ext/rope>
#include <algorithm>
#include <iostream>
using namespace std;
#define ll long long
#define minv 1e-6
#define inf -1e18
#define pi 3.1415926536
#define nl 2.7182818284
const ll mod=1e9+7;//998244353
const int maxn=1e5+10;

ll f[1<<20];

struct node
{
    ll value;
    int pos,t;
};

struct cmp1
{
    bool operator() (node a,node b)
    {
        return a.value>b.value;
    }
};

//priority_queue<node>st;
priority_queue<node,vector<node>,cmp1>st;
int er[21],a[21],b[21],v[21];

int main()
{
    int n,m,s,i,j,value,pos,t;
    ll r=0,rr;
    for (i=0;i<20;i++)
        er[i]=(1<<i);

    scanf("%d",&n);
    for (i=0;i<(1<<n);i++)
            f[i]=inf;

    for (i=0;i<n;i++)
    {
        scanf("%d%d%d",&a[i],&b[i],&m);
        v[i]=0;
        while (m--)
        {
            scanf("%d",&s);
            v[i]+=(er[s-1]);
        }
        if (v[i]==0)
        {
            f[er[i]]=a[i]+b[i];
            st.push({a[i]+b[i],er[i],2});
            r=max(r,(ll)a[i]+b[i]);
        }
    }

    while (!st.empty())
    {
        value=st.top().value;
        pos=st.top().pos;
        t=st.top().t;
        st.pop();
        for (i=0;i<n;i++)
            if ((pos & er[i])==0 && (pos & v[i])==v[i] && value+1ll*t*a[i]+b[i]>f[pos|er[i]])
            {
                rr=value+1ll*t*a[i]+b[i];
                f[pos|er[i]]=rr;
                r=max(r,rr);
                st.push({rr,pos|er[i],t+1});
            }
    }
    printf("%lld",r);
    return 0;
}
/*
3
1 2 0
1 5 0
8 1 0
*/

 

G. Lpl and Energy-saving Lamps

  1 #include <cstdio>
  2 #include <cstdlib>
  3 #include <cmath>
  4 #include <cstring>
  5 #include <time.h>
  6 #include <string>
  7 #include <set>
  8 #include <map>
  9 #include <list>
 10 #include <stack>
 11 #include <queue>
 12 #include <vector>
 13 #include <bitset>
 14 #include <ext/rope>
 15 #include <algorithm>
 16 #include <iostream>
 17 using namespace std;
 18 #define ll long long
 19 #define minv 1e-6
 20 #define inf 1e18
 21 #define pi 3.1415926536
 22 #define nl 2.7182818284
 23 const ll mod=1e9+7;//998244353
 24 const int maxn=1e5+10;
 25 
 26 struct node
 27 {
 28     int x,y;
 29 }f[maxn],r[maxn];
 30 
 31 int tag[maxn<<2],a[maxn],x,y;
 32 
 33 int cmp(node a,node b)
 34 {
 35     return a.x<b.x;
 36 }
 37 
 38 void build(int index,int l,int r)
 39 {
 40     if (l==r)
 41         scanf("%d",&tag[index]);
 42     else
 43     {
 44         int m=(l+r)>>1;
 45         build(index<<1,l,m);
 46         build(index<<1|1,m+1,r);
 47         tag[index]=min(tag[index<<1],tag[index<<1|1]);
 48     }
 49 }
 50 
 51 int query(int index,int l,int r,int v)
 52 {
 53     if (l==r)
 54     {
 55         if (tag[index]>v)
 56             return 0;
 57 
 58         x++;
 59         y-=tag[index];
 60         tag[index]=inf;
 61         return l;
 62     }
 63     else
 64     {
 65         int m=(l+r)>>1,z;
 66         if (tag[index<<1]<=v)
 67             z=query(index<<1,l,m,v);
 68 
 69         else
 70             z=query(index<<1|1,m+1,r,v);
 71         tag[index]=min(tag[index<<1],tag[index<<1|1]);
 72         return z;
 73     }
 74 }
 75 
 76 int main()
 77 {
 78     int n,m,q,Q,d,index,i,j;
 79     scanf("%d%d",&n,&m);
 80     build(1,1,n);
 81 
 82     scanf("%d",&q);
 83     for (Q=1;Q<=q;Q++)
 84     {
 85         scanf("%d",&d);
 86         f[Q].x=d;
 87         f[Q].y=Q;
 88     }
 89     sort(f+1,f+q+1,cmp);
 90 
 91     index=1;
 92     x=0;
 93     y=0;
 94     for (i=1;i<=f[q].x;i++)
 95     {
 96         y+=m;
 97         j=1;
 98         while (j)
 99             j=query(1,1,n,y);
100 
101         while (f[index].x==i)
102         {
103             r[f[index].y]={x,y};
104             index++;
105         }
106     }
107 
108     for (i=1;i<=q;i++)
109         printf("%d %d
",r[i].x,r[i].y);
110     return 0;
111 }

 

J. Sum

 1 #include <cstdio>
 2 #include <cstdlib>
 3 #include <cmath>
 4 #include <cstring>
 5 #include <time.h>
 6 #include <string>
 7 #include <set>
 8 #include <map>
 9 #include <list>
10 #include <stack>
11 #include <queue>
12 #include <vector>
13 #include <bitset>
14 #include <ext/rope>
15 #include <algorithm>
16 #include <iostream>
17 using namespace std;
18 #define ll long long
19 #define minv 1e-6
20 #define inf 1e9
21 #define pi 3.1415926536
22 #define nl 2.7182818284
23 const ll mod=1e9+7;//998244353
24 const int maxn=2e7+10;
25 
26 int zhi[maxn],f[maxn];
27 bool vis[maxn],v[maxn];
28 
29 int main()
30 {
31     int t,i,j,k,x=0,y=0,value=2e7;
32     ll n,sum;
33     memset(vis,0,sizeof(vis));
34     memset(v,0,sizeof(v));
35     for (i=2;i<=value;i++)
36     {
37         if (!vis[i])
38         {
39             x++;
40             zhi[x]=i;
41         }
42         for (j=1;j<=x;j++)
43         {
44             k=i*zhi[j];
45             if (k>value)
46                 break;
47             vis[k]=1;
48             v[k]=v[i];
49             if (i%zhi[j]==0)
50             {
51                 v[k]=1;
52                 break;
53             }
54         }
55     }
56     for (i=1;i<=value;i++)
57         if (!v[i])
58         {
59             y++;
60             f[y]=i;
61         }
62 
63 //    for (i=1;i<=100;i++)
64 //        printf("%d ",f[i]);
65 
66     scanf("%d",&t);
67     f[0]=0;
68     while (t--)
69     {
70         scanf("%lld",&n);
71         sum=0;
72         j=y;
73         for (i=1;i<=y;i++)
74         {
75             while (1ll*f[i]*f[j]>n)
76                 j--;
77             sum+=j;
78         }
79         printf("%lld
",sum);
80     }
81     return 0;
82 }

 

L. Magical Girl Haze

  1 /*
  2 图问题:
  3 spfa容易被卡时间复杂度
  4 
  5 而dijkstra是贪心，不会被卡
  6 */
  7 #include <cstdio>
  8 #include <cstdlib>
  9 #include <cmath>
 10 #include <cstring>
 11 #include <time.h>
 12 #include <string>
 13 #include <set>
 14 #include <map>
 15 #include <list>
 16 #include <stack>
 17 #include <queue>
 18 #include <vector>
 19 #include <bitset>
 20 #include <ext/rope>
 21 #include <algorithm>
 22 #include <iostream>
 23 using namespace std;
 24 #define ll long long
 25 #define minv 1e-6
 26 #define inf 1e18
 27 #define pi 3.1415926536
 28 #define nl 2.7182818284
 29 const ll mod=1e9+7;//998244353
 30 const int maxn=1e5+10;
 31 
 32 struct rec
 33 {
 34     int d,len;
 35     rec *to;
 36 }*e[maxn];
 37 
 38 struct node
 39 {
 40     ll dist;
 41     int k,d;
 42 };
 43 
 44 struct cmp1
 45 {
 46     bool operator() (node a,node b)
 47     {
 48         if (a.dist==b.dist)
 49             return a.k>b.k;
 50         else
 51             return a.dist>b.dist;
 52     }
 53 };
 54 
 55 priority_queue<node,vector<node>,cmp1>st;
 56 
 57 ll f[maxn][12];
 58 
 59 int main()
 60 {
 61     rec *p;
 62     int t,n,m,k,i,j,d,kk,dd,x,y,z;
 63     ll dist,r;
 64     scanf("%d",&t);
 65     while (t--)
 66     {
 67         scanf("%d%d%d",&n,&m,&k);
 68         for (i=1;i<=n;i++)
 69             e[i]=NULL;
 70         while (m--)
 71         {
 72             scanf("%d%d%d",&x,&y,&z);
 73             p=(rec*) malloc (sizeof(rec));
 74             p->d=y;
 75             p->len=z;
 76             p->to=e[x];
 77             e[x]=p;
 78         }
 79         for (i=1;i<=n;i++)
 80             for (j=0;j<=k;j++)
 81                 f[i][j]=inf;
 82         for (j=0;j<=k;j++)
 83         {
 84             f[1][j]=0;
 85             st.push({0,j,1});
 86         }
 87         //f[i][j]作为一个状态，100000*10
 88         while (!st.empty())
 89         {
 90             dist=st.top().dist;
 91             d=st.top().d;
 92             kk=st.top().k;
 93             st.pop();
 94             p=e[d];
 95             while (p)
 96             {
 97                 dd=p->d;
 98                 if (f[dd][kk]>dist+p->len)
 99                 {
100                     f[dd][kk]=dist+p->len;
101                     st.push({f[dd][kk],kk,dd});
102                 }
103                 if (kk!=k && f[dd][kk+1]>dist)
104                 {
105                     f[dd][kk+1]=dist;
106                     st.push({f[dd][kk+1],kk+1,dd});
107                 }
108                 p=p->to;
109             }
110         }
111         r=inf;
112         for (j=0;j<=k;j++)
113             r=min(r,f[n][j]);
114         printf("%lld
",r);
115     }
116     return 0;
117 }