2017中国大学生程序设计竞赛 - 网络选拔赛

Posted 2020-10-04 ISGuXing

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Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6514    Accepted Submission(s): 1610

Problem Description

It is well known that small groups are not conducive of the development of a team. Therefore, there shouldn’t be any small groups in a good team.
In a team with n members,if there are three or more members are not friends with each other or there are three or more members who are friends with each other. The team meeting the above conditions can be called a bad team.Otherwise,the team is a good team.
A company is going to make an assessment of each team in this company. We have known the team with n members and all the friend relationship among these n individuals. Please judge whether it is a good team.

 

Input

The first line of the input gives the number of test cases T; T test cases follow.（T<=15）
The first line od each case should contain one integers n, representing the number of people of the team.(n≤3000 )

Then there are n-1 rows. The i th row should contain n-i numbers, in which number aij represents the relationship between member i and member j+i. 0 means these two individuals are not friends. 1 means these two individuals are friends.

 

Output

Please output ”Great Team!” if this team is a good team, otherwise please output “Bad Team!”.

Sample Input

1

4

1 1 0

0 0

1

Sample Output

Great Team!

题目大意：判断是否某个人有三个及三个以上的朋友或非朋友。

签到题，思路：人数大于5的时候一定是bad team。

模拟就ok。

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<string.h>
 4 #include<algorithm>
 5 using namespace std;
 6 int num[3500];
 7 int main(){
 8     int T;
 9     cin>>T;
10     while(T--){
11         int n;
12         cin>>n;
13         int flag=0;
14         if(n>5) flag=1;
15         int x;
16         memset(num,0,sizeof(num));
17         for(int i=1;i<n;i++){
18             for(int j=1;j<=n-i;j++){
19                 scanf("%d",&x);
20                 if(x==1){
21                     num[i]++;
22                     num[i+j]++;
23                 }
24             }
25         }
26         if(flag) cout<<"Bad Team!"<<endl;
27         else{
28             for(int i=1;i<=n;i++){
29                 if(num[i]>=3||(n-1-num[i])>=3) flag=1;
30             }
31             if(flag) cout<<"Bad Team!"<<endl;
32             else cout<<"Great Team!"<<endl;
33         }
34     }
35     return 0;
36 }

A Secret

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 256000/256000 K (Java/Others)
Total Submission(s): 4199    Accepted Submission(s): 495

Problem Description

Today is the birthday of SF,so VS gives two strings S1,S2 to SF as a present,which have a big secret.SF is interested in this secret and ask VS how to get it.There are the things that VS tell:
  Suffix(S2,i) = S2[i...len].Ni is the times that Suffix(S2,i) occurs in S1 and Li is the length of Suffix(S2,i).Then the secret is the sum of the product of Ni and Li.
  Now SF wants you to help him find the secret.The answer may be very large, so the answer should mod 1000000007.

 

Input

Input contains multiple cases.
  The first line contains an integer T,the number of cases.Then following T cases.
  Each test case contains two lines.The first line contains a string S1.The second line contains a string S2.
  1<=T<=10.1<=|S1|,|S2|<=1e6.S1 and S2 only consist of lowercase ,uppercase letter.

 

Output

For each test case,output a single line containing a integer,the answer of test case.
  The answer may be very large, so the answer should mod 1e9+7.

 

Sample Input

2

aaaaa

aa

abababab

aba

Sample Output

13

19

Hint

case 2: Suffix(S2,1) = "aba", Suffix(S2,2) = "ba", Suffix(S2,3) = "a". N1 = 3, N2 = 3, N3 = 4. L1 = 3, L2 = 2, L3 = 1. ans = （3*3+3*2+4*1）%1000000007.

 

思路：写的时候觉得是kmp但是想了一下可能会超时。没做，后来zll写了超时，后面有人改了，还是用kmp AC了，666啊。

果然还是要写了再说啊

套用一下大佬代码：

 1 #include <iostream> 
 2 #include <algorithm> 
 3 #include <cstring> 
 4 #include <cstdio>
 5 #include <vector> 
 6 #include <queue> 
 7 #include <cstdlib> 
 8 #include <iomanip>
 9 #include <cmath>
10 #include <cassert>
11 #include <ctime> 
12 #include <map> 
13 #include <set> 
14 using namespace std; 
15 #define lowbit(x) (x&(-x)) 
16 #define max(x,y) (x>y?x:y) 
17 #define min(x,y) (x<y?x:y) 
18 #define MAX 100000000000000000 
19 #define MOD 1000000007
20 #define pi acos(-1.0) 
21 #define ei exp(1) 
22 #define PI 3.141592653589793238462
23 #define ios() ios::sync_with_stdio(false)
24 #define INF 1044266558
25 #define mem(a) (memset(a,0,sizeof(a)))
26 typedef long long ll;
27 const int maxn = 1e6+10;
28 #define next HKJournalist
29 ll next[maxn],i;
30 ll val[maxn],t,j,la,lb;
31 char sa[maxn], sb[maxn];
32 void get_next()
33 {
34     next[0]=-1;next[1]=0;
35     val[0]=0;val[1]=1;
36     for(i=1;i<lb;i++)
37     {
38         next[i+1]=0;
39         val[i+1]=i+1;
40         j=next[i];
41         while(j>=0)
42         {
43             if(sb[i]==sb[j])
44             {
45                 next[i+1]=j+1;
46                 val[i+1]+=val[j+1];
47                 val[i+1]%=MOD;
48                 break;
49             }
50             j=next[j];
51         }
52     }
53 }
54 ll find()
55 {
56     ll ans=0;
57     int k=0;
58     for(i=0;i<la;i++)
59     {
60         j=k;
61         for(k=0;j>=0;j=next[j])
62             if(sa[i]==sb[j])
63             {
64                 k=j+1;
65                 break;
66             }
67         ans = (ans+val[k])%MOD;
68     }
69     return ans;
70 }
71 
72 int main ()
73 {
74     scanf("%d", &t);
75     while(t--)
76     {
77         scanf("%s",sa);
78         scanf("%s",sb);
79         la=strlen(sa);
80         lb=strlen(sb);
81         reverse(sa,sa+la);
82         reverse(sb,sb+lb);
83         get_next();
84         printf("%lld\n",find());
85     }
86     return 0;
87 }

 

CaoHaha‘s staff

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2089    Accepted Submission(s): 818

Problem Description

"You shall not pass!"
After shouted out that,the Force Staff appered in CaoHaha‘s hand.
As we all know,the Force Staff is a staff with infinity power.If you can use it skillful,it may help you to do whatever you want.
But now,his new owner,CaoHaha,is a sorcerers apprentice.He can only use that staff to send things to other place.
Today,Dreamwyy come to CaoHaha.Requesting him send a toy to his new girl friend.It was so far that Dreamwyy can only resort to CaoHaha.
The first step to send something is draw a Magic array on a Magic place.The magic place looks like a coordinate system,and each time you can draw a segments either on cell sides or on cell diagonals.In additional,you need 1 minutes to draw a segments.
If you want to send something ,you need to draw a Magic array which is not smaller than the that.You can make it any deformation,so what really matters is the size of the object.
CaoHaha want to help dreamwyy but his time is valuable(to learn to be just like you),so he want to draw least segments.However,because of his bad math,he needs your help.

 

Input

The first line contains one integer T(T<=300).The number of toys.
Then T lines each contains one intetger S.The size of the toy(N<=1e9).

 

Output

Out put T integer in each line ,the least time CaoHaha can send the toy.

 

Sample Input

5

1

2

3

4

5

 

Sample Output

4

4

6

6

7

 

 

题目大意：凑成一个面积为n的图形至少几笔能画好，第五种情况折腾半天画不出。

后来知道了还是找不到规律。。。

还是dalao的代码：

附上一篇好理解的博客：http://blog.csdn.net/qq_36345036/article/details/77415176

 1 #include <bits/stdc++.h>
 2 #define ll long long int
 3 #define N 1000005
 4 using namespace std;
 5 int k[N];
 6 void fa(){
 7     memset(k,0,sizeof(k));
 8     k[0]=k[1]=0;
 9     for(int i=2;i<N;i++){
10         if(i%2==0){
11             k[i]=i/4*2+k[i-2];
12         }else{
13             k[i]=(i+1)/4+k[i-1]-1;
14         }
15     }
16 }
17 int main(){
18     int n;
19     fa();
20     scanf("%d",&n);
21     while(n--){
22         int x;
23         scanf("%d",&x);
24         for(int i=0;i<N;i++){
25             if(k[i]>=x){
26                 printf("%d\n",i);
27                 break;
28             }
29         }
30     }
31     return 0;
32 }