• 2019HPU-ICPC-Training-1

    byl太强了，学弟们太强了～全程被吊打，嘤嘤嘤～

    A题  Connecting Vertices

    http://codeforces.com/problemset/problem/888/F

    不会

    B题 Local Extrema

    http://codeforces.com/problemset/problem/888/A

    给一列数字，判断一个数它的左右是否同时比它大，或者同时比它小，若满足的话那么count++，得到最后的count值，那么很明显，直接暴力的遍历一遍就可以了。

     1 #include <bits/stdc++.h>
 2 #define ll long long
 3 #define ull unsigned long long
 4 #define ms(a,b) memset(a,b,sizeof(a))
 5 const int inf=0x3f3f3f3f;
 6 const ll INF=0x3f3f3f3f3f3f3f3f;
 7 const int maxn=1e6+10;
 8 const int mod=1e9+7;
 9 const int maxm=1e3+10;
10 using namespace std;
11 int a[maxn];
12 int main(int argc, char const *argv[])
13 {
14     #ifndef ONLINE_JUDGE
15         freopen("in.txt", "r", stdin);
16         freopen("out.txt", "w", stdout);
17         srand((unsigned int)time(NULL));
18     #endif
19     ios::sync_with_stdio(false);
20     cin.tie(0);
21     int n;
22     cin>>n;
23     for(int i=1;i<=n;i++)
24         cin>>a[i];
25     int ans=0;
26     for(int i=2;i<n;i++)
27     {
28         if(a[i]<a[i-1]&&a[i]<a[i+1])
29             ans++;
30         if(a[i]>a[i-1]&&a[i]>a[i+1])
31             ans++;
32     }
33     cout<<ans<<endl;
34     #ifndef ONLINE_JUDGE
35         cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
36     #endif
37     return 0;
38 }
    View Code

    C题 Xor-MST

    http://codeforces.com/problemset/problem/888/G

    不会

    D题 Buggy Robot

    http://codeforces.com/problemset/problem/888/B

    机器人有四种指令，找出在给出的一大串指令中，最多有多少指令是正确的

    找到LR和UD有多少对，乘以2就行了

     1 #include <bits/stdc++.h>
 2 #define ll long long
 3 #define ull unsigned long long
 4 #define ms(a,b) memset(a,b,sizeof(a))
 5 const int inf=0x3f3f3f3f;
 6 const ll INF=0x3f3f3f3f3f3f3f3f;
 7 const int maxn=1e6+10;
 8 const int mod=1e9+7;
 9 const int maxm=1e3+10;
10 using namespace std;
11 int main(int argc, char const *argv[])
12 {
13     #ifndef ONLINE_JUDGE
14         freopen("in.txt", "r", stdin);
15         freopen("out.txt", "w", stdout);
16         srand((unsigned int)time(NULL));
17     #endif
18     ios::sync_with_stdio(false);
19     cin.tie(0);
20     int n;
21     cin>>n;
22     string s;
23     cin>>s;
24     map<char,int>mp;
25     for(int i=0;i<n;i++)
26         mp[s[i]]++;
27     int ans=0;
28     ans+=min(mp['L'],mp['R']);
29     ans+=min(mp['U'],mp['D']);
30     cout<<ans*2<<endl;
31     #ifndef ONLINE_JUDGE
32         cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
33     #endif
34     return 0;
35 }
    View Code

    E题 K-Dominant Character

    http://codeforces.com/problemset/problem/888/C

    给出一个字符串，找出一个最小的长度k，使得每个长度为k的子串中都包含一个相同的字符记录下来每个字符的位置，找两个相同字符的最大距离，对这个最大距离取最小值

     1 #include <bits/stdc++.h>
 2 #define ll long long
 3 #define ull unsigned long long
 4 #define ms(a,b) memset(a,b,sizeof(a))
 5 const int inf=0x3f3f3f3f;
 6 const ll INF=0x3f3f3f3f3f3f3f3f;
 7 const int maxn=1e6+10;
 8 const int mod=1e9+7;
 9 const int maxm=1e3+10;
10 using namespace std;
11 int main(int argc, char const *argv[])
12 {
13     #ifndef ONLINE_JUDGE
14         freopen("in.txt", "r", stdin);
15         freopen("out.txt", "w", stdout);
16         srand((unsigned int)time(NULL));
17     #endif
18     ios::sync_with_stdio(false);
19     cin.tie(0);
20     string s;
21     cin>>s;
22     int l=s.length();
23     vector<int>ve[30];
24     for(int i=0;i<26;i++)
25         ve[i].push_back(-1);
26     for(int i=0;i<l;i++)
27         ve[s[i]-'a'].push_back(i);
28     for(int i=0;i<26;i++)
29         ve[i].push_back(l);
30     int ans=inf;
31     for(int i=0;i<26;i++)
32     {
33         int res=0;
34         int sz=ve[i].size();
35         for(int j=1;j<sz-1;j++)
36             res=max(res,max(ve[i][j]-ve[i][j-1],ve[i][j+1]-ve[i][j]));
37         if(res==0)
38             continue;
39         ans=min(ans,res);
40     }
41     cout<<ans<<endl;
42     #ifndef ONLINE_JUDGE
43         cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
44     #endif
45     return 0;
46 }
    View Code

    F题 Maximum Subsequence

    http://codeforces.com/problemset/problem/888/E

    给出n个数，从这n个数中选出几个数（可以不选），使得这些数的和对m取余后的值最大

    题解链接：https://www.cnblogs.com/Friends-A/p/11569017.html

     1 #include <bits/stdc++.h>
 2 #define ll long long
 3 #define ull unsigned long long
 4 #define ms(a,b) memset(a,b,sizeof(a))
 5 const int inf=0x3f3f3f3f;
 6 const ll INF=0x3f3f3f3f3f3f3f3f;
 7 const int maxn=1e6+10;
 8 const int mod=1e9+7;
 9 const int maxm=1e3+10;
10 using namespace std;
11 int a[maxn];
12 int Left[maxn];
13 int Right[maxn];
14 int cntl,cntr;
15 int n,m;
16 int main(int argc, char const *argv[])
17 {
18     #ifndef ONLINE_JUDGE
19         freopen("in.txt", "r", stdin);
20         freopen("out.txt", "w", stdout);
21         srand((unsigned int)time(NULL));
22     #endif
23     ios::sync_with_stdio(false);
24     cin.tie(0);
25     cin>>n>>m;
26     for(int i=0;i<n;i++)
27         cin>>a[i],a[i]%=m;
28     int res=0;
29     int l,r;
30     l=r=n/2;
31     for(int i=0;i<(1<<r);i++)
32     {
33         res=0;
34         for(int j=0;j<r;j++)
35             if(i>>j&1)
36                 res+=a[j],res%=m;
37         Left[cntl++]=res;
38     }
39     res=0;
40     r=n;
41     int num=r-l+1;
42     for(int i=0;i<(1<<num);i++)
43     {
44         res=0;
45         for(int j=0;j<num;j++)
46             if(i>>j&1)
47                 res+=a[l+j],res%=m;
48         Right[cntr++]=res;
49     }
50     Left[cntl++]=0;
51     Right[cntr++]=0;
52     sort(Left,Left+cntl);
53     sort(Right,Right+cntr);
54     cntl=unique(Left,Left+cntl)-Left;
55     cntr=unique(Right,Right+cntr)-Right;
56     int ans=0;
57     for(int i=0;i<cntl;i++)
58     {
59         int res=m-Left[i]-1;
60         int pos=upper_bound(Right,Right+cntr,res)-Right;
61         int num=Right[pos-1];
62         ans=max(ans%m,(num+Left[i])%m);
63     }
64     cout<<ans<<endl;
65     #ifndef ONLINE_JUDGE
66         cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
67     #endif
68     return 0;
69 }
    View Code

    G题 Almost Identity Permutations

    http://codeforces.com/problemset/problem/888/D

    给出n的全排列，求有多少种排列，满足至少n−k个位置上的数和下标相同（下标从1开始）

    错排公式+组合数

    题解链接：https://www.cnblogs.com/Friends-A/p/11569153.html

     1 #include <bits/stdc++.h>
 2 #define ll long long
 3 #define ull unsigned long long
 4 #define ms(a,b) memset(a,b,sizeof(a))
 5 const int inf=0x3f3f3f3f;
 6 const ll INF=0x3f3f3f3f3f3f3f3f;
 7 const int maxn=1e6+10;
 8 const int mod=1e9+7;
 9 const int maxm=1e3+10;
10 using namespace std;
11 ll C(int n,int m)
12 {
13     ll fenmu=1LL;
14     ll fenzi=1LL;
15     for(int i=1;i<=m;i++)
16     {
17         fenmu=1LL*fenmu*(n-i+1);
18         fenzi=1LL*fenzi*i;
19     }
20     return fenmu/fenzi;
21 }
22 int main(int argc, char const *argv[])
23 {
24     #ifndef ONLINE_JUDGE
25         freopen("in.txt", "r", stdin);
26         freopen("out.txt", "w", stdout);
27         srand((unsigned int)time(NULL));
28     #endif
29     ios::sync_with_stdio(false);
30     cin.tie(0);
31     int n,k;
32     cin>>n>>k;
33     ll ans=0;
34     if(k>=1)
35         ans+=1;
36     if(k>=2)
37         ans+=(n*(n-1)/2);
38     if(k>=3)
39         ans+=2*C(n,3);
40     if(k>=4)
41         ans+=9*C(n,4);
42     cout<<ans<<endl;
43     #ifndef ONLINE_JUDGE
44         cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
45     #endif
46     return 0;
47 }
    View Code

    H题 Alyona and Spreadsheet

    http://codeforces.com/problemset/problem/777/C

    给出一个n×m的矩阵，判断第l行～第r行中是否有一列是非递减的

    预处理每一行能往上延伸到的位置，注意矩阵的存法

    题解链接：https://www.cnblogs.com/Friends-A/p/11569247.html

     1 #include <bits/stdc++.h>
 2 #define ll long long
 3 #define ull unsigned long long
 4 #define ms(a,b) memset(a,b,sizeof(a))
 5 const int inf=0x3f3f3f3f;
 6 const ll INF=0x3f3f3f3f3f3f3f3f;
 7 const int maxn=1e6+10;
 8 const int mod=1e9+7;
 9 const int maxm=1e3+10;
10 using namespace std;
11 vector<int>ve[maxn];
12 // 当前行能往上延伸的最高位置
13 int can[maxn];
14 // 当前列能往上的最高位置
15 int line[maxn];
16 int main(int argc, char const *argv[])
17 {
18     #ifndef ONLINE_JUDGE
19         freopen("in.txt", "r", stdin);
20         freopen("out.txt", "w", stdout);
21         srand((unsigned int)time(NULL));
22     #endif
23     ios::sync_with_stdio(false);
24     cin.tie(0);
25     int n,m;
26     cin>>n>>m;
27     int x;
28     for(int i=0;i<m;i++)
29         ve[0].push_back(0);
30     for(int i=1;i<=n;i++)
31         for(int j=0;j<m;j++)
32             cin>>x,ve[i].push_back(x);
33     for(int i=1;i<=n;i++)
34     {
35         can[i]=i;
36         for(int j=0;j<m;j++)
37         {
38             int now_num=ve[i][j];
39             int up_num=ve[i-1][j];
40             if(now_num<up_num)
41                 line[j]=i;
42             can[i]=min(can[i],line[j]);
43         }
44     }
45     int t;
46     cin>>t;
47     while(t--)
48     {
49         int l,r;
50         cin>>l>>r;
51         if(can[r]>l)
52             cout<<"No
";
53         else
54             cout<<"Yes
";
55     }
56     #ifndef ONLINE_JUDGE
57         cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
58     #endif
59     return 0;
60 }
    View Code

    I题 Shell Game

    http://codeforces.com/problemset/problem/777/A

    现在一共有三个小盒子，其中有一个盒子中有小球.一共进行了n次操作，操作规律有：

    ①奇数次操作，交换第一个和中间的盒子。

    ②偶数次操作，交换第三个和中间的盒子。

    现在已知操作了n次之后小球在x号盒子中（0,1,2），问初始的时候小球在哪里

    循环节，每六个数字一个循环

     1 #include <bits/stdc++.h>
 2 #define ll long long
 3 #define ull unsigned long long
 4 #define ms(a,b) memset(a,b,sizeof(a))
 5 const int inf=0x3f3f3f3f;
 6 const ll INF=0x3f3f3f3f3f3f3f3f;
 7 const int maxn=1e6+10;
 8 const int mod=1e9+7;
 9 const int maxm=1e3+10;
10 using namespace std;
11 int main(int argc, char const *argv[])
12 {
13     #ifndef ONLINE_JUDGE
14         freopen("in.txt", "r", stdin);
15         freopen("out.txt", "w", stdout);
16         srand((unsigned int)time(NULL));
17     #endif
18     ios::sync_with_stdio(false);
19     cin.tie(0);
20     int a[6][3]={{0,1,2},{1,0,2},{1,2,0},{2,1,0},{2,0,1},{0,2,1}};
21     int n,x;
22     cin>>n>>x;
23     n%=6;
24     cout<<a[n][x]<<endl;
25     #ifndef ONLINE_JUDGE
26         cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
27     #endif
28     return 0;
29 }
    View Code

    J题 Hanoi Factory

    http://codeforces.com/problemset/problem/777/E

    有n个空心圆柱体，第i个圆柱体的内径、外径、高分别为：ai,bi,hi。将这些圆柱体堆起来，要求：从上到下，外径非递减，并且上面的外径小于下面的内径。问最高能堆多高

    贪心，用栈维护

    题解链接：https://www.cnblogs.com/Friends-A/p/11571769.html

     1 #include <bits/stdc++.h>
 2 #define ll long long
 3 #define ull unsigned long long
 4 #define ms(a,b) memset(a,b,sizeof(a))
 5 const int inf=0x3f3f3f3f;
 6 const ll INF=0x3f3f3f3f3f3f3f3f;
 7 const int maxn=1e6+10;
 8 const int mod=1e9+7;
 9 const int maxm=1e3+10;
10 using namespace std;
11 struct wzy
12 {
13     int a,b,h;
14 }p[maxn];
15 bool cmp(wzy u,wzy v)
16 {
17     if(u.b==v.b)
18     {
19         if(u.a==v.a)
20             return u.h>v.h;
21         return u.a>v.a;
22     }
23     return u.b>v.b;
24 }
25 int main(int argc, char const *argv[])
26 {
27     #ifndef ONLINE_JUDGE
28         freopen("/home/wzy/in.txt", "r", stdin);
29         freopen("/home/wzy/out.txt", "w", stdout);
30         srand((unsigned int)time(NULL));
31     #endif
32     ios::sync_with_stdio(false);
33     cin.tie(0);
34     int n;
35     cin>>n;
36     for(int i=1;i<=n;i++)
37         cin>>p[i].a>>p[i].b>>p[i].h;
38     sort(p+1,p+1+n,cmp);
39     ll ans=1LL*p[1].h;
40     ll sum=1LL*p[1].h;
41     stack<wzy>st;
42     st.push(p[1]);
43     for(int i=2;i<=n;i++)    
44     {
45         while(!st.empty()&&(st.top().a>=p[i].b||st.top().b<p[i].b))
46         {
47             sum-=1LL*st.top().h;
48             st.pop();
49         }
50         sum+=1LL*p[i].h;
51         st.push(p[i]);
52         ans=max(ans,sum);
53     }
54     cout<<ans<<endl;
55     #ifndef ONLINE_JUDGE
56         cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
57     #endif
58     return 0;
59 }
    View Code

    K题 Cloud of Hashtags

     http://codeforces.com/contest/777/problem/D

    n个字符串，要求不改变位置，删除最少的字符串的后缀，使这些字符串按照字典序非递减的顺序排列

    暴力即可

    题解链接：https://www.cnblogs.com/Friends-A/p/11569328.html

     1 #include <bits/stdc++.h>
 2 #define ll long long
 3 #define ull unsigned long long
 4 #define ms(a,b) memset(a,b,sizeof(a))
 5 const int inf=0x3f3f3f3f;
 6 const ll INF=0x3f3f3f3f3f3f3f3f;
 7 const int maxn=1e6+10;
 8 const int mod=1e9+7;
 9 const int maxm=1e3+10;
10 using namespace std;
11 vector<string>ve;
12 vector<string>ans;
13 int get_place(string s1,string s2)
14 {
15     int l1=s1.length();
16     int l2=s2.length();
17     int i;
18     for(i=0;i<min(l2,l1);i++)
19     {
20         if(s1[i]<s2[i])
21             return l1;
22         if(s1[i]==s2[i])
23             continue;
24         if(s1[i]>s2[i])
25             return i;
26     }
27     if(i==l2)
28     {
29         if(l1>l2)
30         {
31             if(s1[i-1]==s2[i-1])
32                 return l2;
33             else
34                 return l1;
35         }
36     }
37     return l1;
38 }
39 int main(int argc, char const *argv[])
40 {
41     #ifndef ONLINE_JUDGE
42         freopen("in.txt", "r", stdin);
43         freopen("out.txt", "w", stdout);
44         srand((unsigned int)time(NULL));
45     #endif
46     ios::sync_with_stdio(false);
47     cin.tie(0);
48     int n;
49     cin>>n;
50     string s;
51     for(int i=0;i<n;i++)
52     {
53         cin>>s;
54         ve.push_back(s);
55     }
56     string s1,s2;
57     ans.push_back(ve[n-1]);
58     for(int i=n-2;i>=0;i--)
59     {
60         s1=ve[i];
61         s2=ans[n-(i+2)];
62         int pos=get_place(s1,s2);
63         string ss;
64         ss=s1.substr(0,pos);
65         ans.push_back(ss);
66     }
67     for(int i=n-1;i>0;i--)
68         cout<<ans[i]<<endl;
69     cout<<ve[n-1]<<endl;
70     #ifndef ONLINE_JUDGE
71         cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
72     #endif
73     return 0;
74 }
    View Code

    L题 Game of Credit Cards

    http://codeforces.com/contest/777/problem/B 

    Sherlock和Moriarty有n张卡片，每个卡片上有一个数字，现在有Sherlock和Moriarty 两个人在比较这些卡片上的数字大小，小的数字需要接受惩罚，Sherlock的卡片顺序是固定的，Moriarty的卡片顺序可以随意变动，求Moriarty的最小接受惩罚次数是多少，Sherlock最大惩罚对方的次数是多少

    排序比较即可

    题解链接：https://www.cnblogs.com/Friends-A/p/11569436.html

     1 #include <bits/stdc++.h>
 2 #define ll long long
 3 #define ull unsigned long long
 4 #define ms(a,b) memset(a,b,sizeof(a))
 5 const int inf=0x3f3f3f3f;
 6 const ll INF=0x3f3f3f3f3f3f3f3f;
 7 const int maxn=1e6+10;
 8 const int mod=1e9+7;
 9 const int maxm=1e3+10;
10 using namespace std;
11 int s[maxn],m[maxn];
12 int nums[100],numm[100];
13 int main(int argc, char const *argv[])
14 {
15     #ifndef ONLINE_JUDGE
16         freopen("in.txt", "r", stdin);
17         freopen("out.txt", "w", stdout);
18         srand((unsigned int)time(NULL));
19     #endif
20     ios::sync_with_stdio(false);
21     cin.tie(0);
22     int n;
23     string s1,s2;
24     cin>>n;
25     cin>>s1>>s2;
26     for(int i=0;i<n;i++)
27         s[i]=s1[i]-'0',m[i]=s2[i]-'0';
28     sort(s,s+n);
29     sort(m,m+n);
30     int pos1=0;
31     int pos2=0;
32     int res1=0;
33     int res2=0;
34     for(int i=0;i<n;i++)
35     {
36         if(pos1>=n&&pos2>=n)
37             break;
38         while(pos1<n&&m[pos1]<s[i])
39             pos1++;
40         while(pos2<n&&m[pos2]<=s[i])
41             pos2++;
42         if(pos1<n)
43             res1++,pos1++;
44         if(pos2<n)
45             res2++,pos2++;
46     }
47     cout<<n-res1<<endl;
48     cout<<res2<<endl;
49     #ifndef ONLINE_JUDGE
50         cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
51     #endif
52     return 0;
53 }
    View Code

     
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  • 原文地址：https://www.cnblogs.com/Friends-A/p/11569559.html
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