problem A

題意

Let LCM(x,y) be the minimum positive integer that is divisible by both x and y. For example, LCM(13,37)=481, LCM(9,6)=18.

You are given two integers l and r. Find two integers x and y such that l≤x<y≤r and l≤LCM(x,y)≤r.

Input
The first line contains one integer t (1≤t≤10000) — the number of test cases.

Each test case is represented by one line containing two integers l and r (1≤l<r≤109).

Output
For each test case, print two integers:

if it is impossible to find integers x and y meeting the constraints in the statement, print two integers equal to −1;
otherwise, print the values of x and y (if there are multiple valid answers, you may print any of them).

思路

貪心就好了，我們可以知道l*2是最小的我們可以找到的可以和l組成最小公倍數的數字，我們直接找這個數字就好了。

problem B

題意

You are given an array a1,a2,…,an, consisting of n positive integers.

Initially you are standing at index 1 and have a score equal to a1. You can perform two kinds of moves:

move right — go from your current index x to x+1 and add ax+1 to your score. This move can only be performed if x<n.
move left — go from your current index x to x−1 and add ax−1 to your score. This move can only be performed if x>1. Also, you can't perform two or more moves to the left in a row.
You want to perform exactly k moves. Also, there should be no more than z moves to the left among them.

What is the maximum score you can achieve?

Input
The first line contains a single integer t (1≤t≤104) — the number of testcases.

The first line of each testcase contains three integers n,k and z (2≤n≤105, 1≤k≤n−1, 0≤z≤min(5,k)) — the number of elements in the array, the total number of moves you should perform and the maximum number of moves to the left you can perform.

The second line of each testcase contains n integers a1,a2,…,an (1≤ai≤104) — the given array.

The sum of n over all testcases does not exceed 3⋅105.

Output
Print t integers — for each testcase output the maximum score you can achieve if you make exactly k moves in total, no more than z of them are to the left and there are no two or more moves to the left in a row.

思路

由於我們不能連續往左走，所以說我們最優的策略一定是在這組資料的相鄰兩項裡面來回遊走。那麼我們需要記錄的就是相鄰兩項的和，並且記錄一個字首和表示不往回走的值，然後我們找到小於右端點的最大的相鄰兩數，然後加上最多可以走的次數，和右端點的字首和取一個max就是答案了。

程式碼實現

#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define rep(i,f_start,f_end) for (int i=f_start;i<=f_end;++i)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define MT(x,i) memset(x,i,sizeof(x) )
#define rev(i,start,end) for (int i=start;i<end;i++)
#define inf 0x3f3f3f3f
#define mp(x,y) make_pair(x,y)
#define lowbit(x) (x&-x)
#define MOD 1000000007
#define exp 1e-8
#define N 1000005 
#define fi first 
#define se second
#define pb push_back
typedef long long ll;
typedef vector <int> VI;
typedef pair<int ,int> PII;
typedef pair<int ,PII> PIII;
ll gcd (ll a,ll b) {return b?gcd (b,a%b):a; }
inline int read() {
    char ch=getchar(); int x=0, f=1;
    while(ch<'0'||ch>'9') {
        if(ch=='-') f = -1;
        ch=getchar();
    } 
    while('0'<=ch&&ch<='9') {
        x=x*10+ch-'0';
        ch=getchar();
    }   return x*f;
}

const int maxn=1e5+10;
int a[maxn];
int sum[maxn];
ll v[maxn];
int t;
int n,k,z;

int main() {
    cin>>t;
    while (t--) {
        scanf ("%d%d%d",&n,&k,&z);
        rep (i,1,n) {
            scanf ("%d",&a[i]);
            sum[i]=sum[i-1]+a[i];
        }
        rev (i,1,n) v[i]=a[i]+a[i+1];

        ll ans=sum[k+1],ed=k+1,m=-1;
        rep (j,1,z) {
            ed=k+1-2*j;
            m=-1;
            if (ed<1) break;
            rep (i,1,ed) m=max (m,v[i]);
            ans=max (sum[ed]+m*j,ans);
        }
        cout<<ans<<endl;
    }
    return 0;
}

problem C

題意

Let's call left cyclic shift of some string t1t2t3…tn−1tn as string t2t3…tn−1tnt1.

Analogically, let's call right cyclic shift of string t as string tnt1t2t3…tn−1.

Let's say string t is good if its left cyclic shift is equal to its right cyclic shift.

You are given string s which consists of digits 0–9.

What is the minimum number of characters you need to erase from s to make it good?

Input
The first line contains single integer t (1≤t≤1000) — the number of test cases.

Next t lines contains test cases — one per line. The first and only line of each test case contains string s (2≤|s|≤2⋅105). Each character si is digit 0–9.

It's guaranteed that the total length of strings doesn't exceed 2⋅105.

Output
For each test case, print the minimum number of characters you need to erase from s to make it goo

思路

首先，我們簡單分析就可以得出，題目所說的牛逼串肯定是隻有一種字母或者兩種字母組成，超出兩種那麼無論如何都無法組成牛逼串，那麼兩種的話一定是穿插組成。那麼我們記錄每個字母出現的位置和次數，暴力列舉就好了。

程式碼實現

#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define rep(i,f_start,f_end) for (int i=f_start;i<=f_end;++i)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define MT(x,i) memset(x,i,sizeof(x) )
#define rev(i,start,end) for (int i=start;i<end;i++)
#define inf 0x3f3f3f3f
#define mp(x,y) make_pair(x,y)
#define lowbit(x) (x&-x)
#define MOD 1000000007
#define exp 1e-8
#define N 1000005 
#define fi first 
#define se second
#define pb push_back
typedef long long ll;
typedef vector <int> VI;
typedef pair<int ,int> PII;
typedef pair<int ,PII> PIII;
ll gcd (ll a,ll b) {return b?gcd (b,a%b):a; }
inline int read() {
    char ch=getchar(); int x=0, f=1;
    while(ch<'0'||ch>'9') {
        if(ch=='-') f = -1;
        ch=getchar();
    } 
    while('0'<=ch&&ch<='9') {
        x=x*10+ch-'0';
        ch=getchar();
    }   return x*f;
}

const int maxn=2e5+10;
int t;

int main() {
    cin>>t;
    while (t--) {
      string s;
      cin>>s;
      vector <VI> pos(10);
      rev (i,0,s.length ()) {
          int x=s[i]-'0';
          pos[x].pb (i);
      }
      int ans=0;
      rev (i,0,10) ans=max (ans, (int) pos[i].size ());
      rev (i,0,10) {
          rev (j,0,10) {
              if (i==j) continue;
              int cnt=0;
              int l=0,r=0,pre=-1;
              while (l<pos[i].size()&&r<pos[j].size ()) {
                  if (pos[i][l]<pos[j][r]&&pos[i][l]>pre) {
                      cnt+=2;
                      pre=pos[j][r];
                      l++,r++;
                  }
                  else if (pos[i][l]>pos[j][r]) r++;
                  else ++l;
              }
              ans=max (ans,cnt);
          }
       }     
       cout<<(int ) s.length ()-ans<<endl;
    }
    return 0;
}