冯志远0912

Posted 2020-10-07

 

 

 

 

 个人感受：这次分数是50+40+0，也就是说两天分数相加是340分，是达不了标的，这次第一题打好以后，想了许久第二题，发现了有重复元素的排列

但是两个元素，就是求一个组合（这个没想到，就算会Lucas也没什么用），╮(╯▽╰)╭，然后第三题，暴力都写不出什么，拿了个0分。

gemo.cpp

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cmath>
 5 #include<cstring>
 6 
 7 using namespace std;
 8 
 9 int n,m,p;
10 int a[100007],b[100007];
11 
12 int main()
13 {
14     freopen("gemo.in","r",stdin);
15     freopen("gemo.out","w",stdout);
16     
17     scanf("%d%d%d",&n,&m,&p);
18     for (int i=1;i<=m;i++)
19         scanf("%d%d",&a[i],&b[i]);
20     printf("%d",n-1-1-p);
21 }

liqi.cpp

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cmath>
 5 #include<cstring>
 6 #include<queue>
 7 #define LL long long
 8 #define NN 100007
 9 #define fzy pair<LL,LL>
10 
11 using namespace std;
12 
13 int n,m;
14 LL len;
15 struct Node
16 {
17     LL l,r;
18 }a[NN];
19 
20 bool solve(int num)
21 {
22     priority_queue<fzy,vector<fzy>,greater<fzy> >q;
23     for (int i=1;i<=num;i++) q.push(make_pair(a[i].l,a[i].r));
24     LL res=0,mx=0;
25     while (!q.empty())
26     {
27         fzy now=q.top();
28         q.pop();
29         if (now.first>mx+1)
30         {
31             int t=now.first-1-mx;
32             if (t%len==0) res+=t/len;
33             else
34             {
35                 res=res+t/len+1;
36                 t=(len-t%len)+now.first-1;
37                 if (t>now.second)
38                     q.push(make_pair(now.first,t));
39                 mx=t;
40             }
41             mx=max(mx,now.second);
42         }
43         else mx=max(mx,now.second);
44     }
45     int t;
46     if (mx<n)
47     {
48         if ((n-mx)%len==0) res+=(n-mx)/len;
49         else res+=(n-mx)/len+1;
50     }
51     if (res<=num) return true;
52     else return false;
53 }
54 int main()
55 {
56     freopen("liqi.in","r",stdin);
57     freopen("liqi.out","w",stdout);
58     
59     scanf("%d%d%lld",&n,&m,&len);
60     for (int i=1;i<=m;i++)
61         scanf("%lld%lld",&a[i].l,&a[i].r);
62     
63     int l=1,r=m;
64     while (l<r)
65     {
66         int mid=(l+r)>>1;
67         if (solve(mid)) r=mid;
68         else l=mid+1;
69     }
70     
71     printf("%d\\n",l);
72 }

zhexue.cpp

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<algorithm>
 6 #define mod 1000003
 7 #define NN 5007
 8 
 9 using namespace std;
10 
11 int n,m,p;
12 int f[NN][NN];
13 long long q[100007];
14 
15 long long ksm(long long a,int b)
16 {
17     long long ans=1;
18     while (b)
19     {
20         if (b&1) ans=ans*a%mod;
21         a=(a*a)%mod;
22         b=b>>1;
23     }
24     return ans;
25 }
26 void solve()
27 {
28     n++,m++;
29     if (n>m)
30     {
31         printf("0\\n");
32         return;
33     }
34     if (n==m)
35     {
36         printf("1\\n");
37         return;
38     }
39     q[1]=1;
40     for (int i=2;i<=m;i++)
41         q[i]=q[i-1]*i%mod;
42     long long x=q[m-1],y=q[n-1]*q[m-n]%mod;
43     y=ksm(y,mod-2);
44     long long ans=(x*y)%mod;
45     printf("%lld\\n",ans);    
46 }
47 int main()
48 {
49     freopen("zhexue.in","r",stdin);
50     freopen("zhexue.out","w",stdout);
51     
52     scanf("%d%d%d",&n,&m,&p);
53     if (n<=5000&&m<=5000)
54     {
55         int x,y;
56         for (int i=1;i<=p;i++)
57         {
58             scanf("%d%d",&x,&y);
59             f[x][y]=-1;
60         }
61         for (int i=1;i<=m;i++)
62             f[1][i]=f[1][i-1]+1;
63         for (int i=2;i<=n;i++)
64             for (int j=i;j<=m;j++)
65                 if (f[i][j]!=-1) f[i][j]=(f[i][j-1]+f[i-1][j-1])%mod;
66                 else f[i][j]=f[i][j-1];
67         printf("%d\\n",f[n][m]);
68     }
69     else if (p==0) solve();
70 }