昨晚因?yàn)橛悬c(diǎn)事就去忙了，沒(méi)打后悔啊

A - XXFESTIVAL

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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.

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Input

Input is given from Standard Input in the following format:

S

Output

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

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Sample Input 1

Copy CODEFESTIVAL

Sample Output 1

Copy CODE

This is the same as the example in the statement.

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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.

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Sample Input 3

Copy YAKINIKUFESTIVAL

Sample Output 3

Copy YAKINIKU

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?

#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

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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.

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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.

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Sample Input 1

Copy 5 3 1 4 1 5 3 5 4 3

Sample Output 1

Copy YES

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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.

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Sample Input 3

Copy 1 800 5 100 100 100 100 100

Sample Output 3

Copy NO

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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

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用map看一下數(shù)據(jù)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

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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.

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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.

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Sample Input 1

Copy 6 5 1 2 2 3 3 4 4 5 5 6

Sample Output 1

Copy 4

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

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Sample Input 2

Copy 5 5 1 2 2 3 3 1 5 4 5 1

Sample Output 2

Copy 5

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.

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搞成二分圖然后dfs啊

然后再利用六哥的兩邊的個(gè)數(shù)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; }

?

轉(zhuǎn)載于:https://www.cnblogs.com/BobHuang/p/7639623.html

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