CODE FESTIVAL 2017 qual B

昨晚因为有点事就去忙了，没打后悔啊

A - XXFESTIVAL

Time limit : 2sec / Memory limit : 256MB

Score : 100 points

Problem Statement

Rng is going to a festival.

The name of the festival is given to you as a string S, which ends with FESTIVAL, from input. Answer the question: "Rng is going to a festival of what?" Output the answer.

Here, assume that the name of "a festival of s" is a string obtained by appending FESTIVAL to the end of s. For example, CODEFESTIVAL is a festival of CODE.

Constraints

9≤|S|≤50
S consists of uppercase English letters.
S ends with FESTIVAL.

Input

Input is given from Standard Input in the following format:

Output

Print the answer to the question: "Rng is going to a festival of what?"

Sample Input 1

Copy

CODEFESTIVAL

Sample Output 1

Copy

CODE

This is the same as the example in the statement.

Sample Input 2

Copy

CODEFESTIVALFESTIVAL

Sample Output 2

Copy

CODEFESTIVAL

This string is obtained by appending FESTIVAL to the end of CODEFESTIVAL, so it is a festival of CODEFESTIVAL.

Sample Input 3

Copy

YAKINIKUFESTIVAL

Sample Output 3

Copy

YAKINIKU

#include<bits/stdc++.h>
using namespace std;
int main()
{string s;cin>>s;int l=s.length()-8;for(int i=0; i<l; ++i)putchar(s[i]);return 0;
}

B - Problem Set

Time limit : 2sec / Memory limit : 256MB

Score : 200 points

Problem Statement

Rng is preparing a problem set for a qualification round of CODEFESTIVAL.

He has N candidates of problems. The difficulty of the i-th candidate is Di.

There must be M problems in the problem set, and the difficulty of the i-th problem must be Ti. Here, one candidate of a problem cannot be used as multiple problems.

Determine whether Rng can complete the problem set without creating new candidates of problems.

Constraints

1≤N≤200,000
1≤Di≤109
1≤M≤200,000
1≤Ti≤109
All numbers in the input are integers.

Partial Score

100 points will be awarded for passing the test set satisfying N≤100 and M≤100.

Input

Input is given from Standard Input in the following format:

N
D1 D2 … DN
M
T1 T2 … TM

Output

Print YES if Rng can complete the problem set without creating new candidates of problems; print NO if he cannot.

Sample Input 1

Copy

Sample Output 1

Copy

YES

Sample Input 2

Copy

7
100 200 500 700 1200 1600 2000
6
100 200 500 700 1600 1600

Sample Output 2

Copy

NO

Not enough 1600s.

Sample Input 3

Copy

1
800
5
100 100 100 100 100

Sample Output 3

Copy

NO

Sample Input 4

Copy

15
1 2 2 3 3 3 4 4 4 4 5 5 5 5 5
9
5 4 3 2 1 2 3 4 5

Sample Output 4

Copy

YES

用map看一下数据是否合法就行了

#include<bits/stdc++.h>
using namespace std;
map<int,int>M;
int main()
{int n,m;cin>>n;for(int i=0; i<n; i++){int x;cin>>x,M[x]++;}cin>>m;int f=0;for(int i=0; i<m; i++){int x;cin>>x;if(!M[x]){f=1;break;}else M[x]--;}cout<<(f?"NO\n":"YES\n");return 0;
}

C - 3 Steps

Time limit : 2sec / Memory limit : 256MB

Score : 500 points

Problem Statement

Rng has a connected undirected graph with N vertices. Currently, there are M edges in the graph, and the i-th edge connects Vertices Ai and Bi.

Rng will add new edges to the graph by repeating the following operation:

Operation: Choose u and v (u≠v) such that Vertex v can be reached by traversing exactly three edges from Vertex u, and add an edge connecting Vertices uand v. It is not allowed to add an edge if there is already an edge connecting Vertices u and v.

Find the maximum possible number of edges that can be added.

Constraints

2≤N≤105
1≤M≤105
1≤Ai,Bi≤N
The graph has no self-loops or multiple edges.
The graph is connected.

Input

Input is given from Standard Input in the following format:

N M
A1 B1
A2 B2
:
AM BM

Output

Find the maximum possible number of edges that can be added.

Sample Input 1

Copy

Sample Output 1

Copy

If we add edges as shown below, four edges can be added, and no more.

Sample Input 2

Copy

Sample Output 2

Copy

Five edges can be added, for example, as follows:

Add an edge connecting Vertex 5 and Vertex 3.
Add an edge connecting Vertex 5 and Vertex 2.
Add an edge connecting Vertex 4 and Vertex 1.
Add an edge connecting Vertex 4 and Vertex 2.
Add an edge connecting Vertex 4 and Vertex 3.

搞成二分图然后dfs啊

然后再利用六哥的两边的个数想乘就好了

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
bool f=0;
int c[N],cnt[3];
vector<int>G[N];
void dfs(int x,int ff=1)
{if(c[x]){f|=ff!=c[x];return;}c[x]=ff;cnt[ff]++;for(int u:G[x])dfs(u,3-ff);
}
int main()
{int n,m;scanf("%d%d",&n,&m);for(int i=0; i<m; ++i){int u,v;scanf("%d%d",&u,&v);G[u].push_back(v);G[v].push_back(u);}dfs(1);printf("%lld\n",f?1LL*n*(n-1)/2-m:1LL*cnt[1]*cnt[2]-m);return 0;
}

转载于:https://www.cnblogs.com/BobHuang/p/7639623.html