2017中国大学生程序设计竞赛 - 网络选拔赛)

 

Friend-Graph

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3254    Accepted Submission(s): 523

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 ith 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!

 题意：有一个表示朋友关系的图，如果有三个或超过三个的人彼此之间都不是朋友

或者有三个或超过三个以上的朋友彼此之间都是朋友，那么这是一个Bad Team，否则是Great Team！

 当n>5时无论怎么连线都是 Bad Team,

而n=1和n=2肯定是Great Team，所以需判断的范围就很小了，就剩下n为3,4，的情况

 

#include <bits/stdc++.h>
using namespace std;
#define maxn 3007
int ans[maxn];

int main(){
    int T,n,t;
    int maxx,minn;
    scanf("%d",&T);
    while(T--){
    maxx=0,minn=maxn;
    memset(ans,0,sizeof(ans));
    scanf("%d",&n);
    for(int i=1;i<n;i++){
      for(int j=i+1;j<=n;j++){
        scanf("%d",&t);
        if(t==1){
        ans[i]++,ans[j]++; //记录图的度数
        maxx=max(maxx,ans[i]);  //保存最大度数
        maxx=max(maxx,ans[j]);
        }
      }
    } 
    for(int i=1;i<=n;i++)
        minn=min(minn,ans[i]);
    if(n>5) printf("Bad Team!\n");
    else if(n<=2) printf("Great Team!\n");
    else if(n==3){   //n为3
        if(minn==maxx&&(maxx=0||maxx==2))  //3个人都是朋友，3个人都不是朋友
            printf("Bad Team!\n");
        else printf("Great Team!\n");
    } 
    else {
        if(maxx<3&&minn>=n-3)    //n为4，画一下图就出来了
            printf("Great Team!\n");
        else printf("Bad Team!\n");
        }
    }
    return 0;
}

 

 

A Secret

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 256000/256000 K (Java/Others)
Total Submission(s): 195    Accepted Submission(s): 83

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.

 

#include <cstdio>
#include <iostream>
#include <string.h>
using namespace std;
#define ll long long 
const int maxn=1e6+5;
int n,m,Next[maxn];
char s[maxn],t[maxn];
ll id[maxn];
const int mod=1e9+7;
void getNext(){
    int i=0,j=-1;
    Next[0]=-1;
        while(i<m){
        if(j==-1||t[i]==t[j])
            Next[++i]=++j;
        else j=Next[j];
        }
}
void kmp(){  
    int i=0,j=0;
    while(i<n){ 
    if(j==-1||s[i]==t[j])
        ++i,++j;
    else j=Next[j]; 
    id[j]++; 
    if(j==m) j=Next[j]; 
    } 
}
int main(){ 
    int T;
    scanf("%d",&T);
    while(T--){
    memset(id,0,sizeof(id));
    scanf("%s%s",s,t);
    n=strlen(s),m=strlen(t);
    for(int i=0;i<=(n-1)>>1;i++)
        swap(s[i],s[n-1-i]);
    for(int i=0;i<=(m-1)>>1;i++)
        swap(t[i],t[m-1-i]);
    getNext();
    kmp();
    ll ans=0;
    for(int i=m;i>0;--i){
    id[Next[i]]+=id[i];
    ans=(ans+id[i]*i)%mod;
        }
    printf("%lld\n",ans);
    }
    return 0;
}