Codeforce codecraft-20 div2

# A.Grade Allocation

# The meaning of problems

Maximum value sign, a given number n, m is the maximum value, the average must remain unchanged, the demand can be achieved

# Explanations

ans=min(m,sum)

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 #define pii pair<int,int>
 4 #define pll pair<ll,ll>
 5 #define fi first
 6 #define se second
 7 #define pb push_back
 8 using namespace std;
 9 const int N=1e5+10;
10 int a[N];
11 void solve(){
12     int n,m;
13     cin>>n>>m;
14     int sum=0;
15     for(int i=1;i<=n;i++){
16         cin>>a[i];
17         sum+=a[i];
18     }
19     if(sum>=m)
20         cout<<m<<endl;
21     else{
22         cout<<sum<<endl;
23     }
24 }
25 int main(){
26     int t;
27     cin>>t;
28     while(t--){
29         solve();
30     }
31 
32 }

# B.String Modification

# The meaning of problems

A string operation may be performed is a number k (1 <= k <= n), i from 1 to n-k 1, the s [i ~ i k-1 +] all reverse +, calculated a k after the string so that the operation is lexicographically smallest, if there is more satisfied k, find the least k

# Explanations

First we have a string s,
we can express it as s1s2s3 ... sn, and we choose a k:
after the first reverse: sksk-1sk-2 ... s1sk + 1sk + 2 ... sn
position is then observed sk would not have happened it changed.
Subjected to a second reverse: sksk + 1s1 ... sk-2sk -1sk + 2 ... sksk + 1 position will not change after observing the sn.
If the third reverse: sksk + 1sk + 2 ... s2s1sk + 3 ... sksk + position 1sk + 2 does not change after re-sn observed.
......
and so on, after the end of the string is reversed, s' n-k + 1 before the characters must be sksk + 1sk + 2 ... sn
may be found by observing, s k-1 in the previous character s1s2s3 ... sk-1 will be moved to s', the latter half, but the direction uncertain.
Reversal number is n-k + 1, when an odd number of inversions is the need to reverse the last splice, the splice even number directly

The time complexity of O (N ² )

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int T,n,ans;
 4 string s,mn;
 5 string work(int k)
 6 {
 7     if(k==n)
 8     {
 9         string res=s;
10         reverse(res.begin(),res.end());
11         return res;
12     }
13     int t=n-k+1;
14     if(t%2==0)
15     {
16         string res=s.substr(k-1,n-k+1);
17         res+=s.substr(0,k-1);
18         return res;
19     }
20     string res=s.substr(k-1,n-k+1);
21     string tmp=s.substr(0,k-1);
22     reverse(tmp.begin(),tmp.end());
23     res=res+tmp;
24     return res;
25 }
26 int main()
27 {
28     scanf("%d",&T);
29     while(T--)
30     {
31         scanf("%d",&n);
32         cin>>s;
33         ans=1,mn=s;
34         for(int i=2;i<=n;i++)
35             if(work(i)<mn)
36                 mn=work(i),ans=i;
37         cout<<mn<<endl<<ans<<endl;
38     }
39 }

 

# C

# The meaning of problems

# answer

# D

# The meaning of problems

# answer

# E