[2016-04-17] [Gym] [100947] C [Rotate It!]

• Time: 2016-04-17-16:35:01 Sunday
• Title Number: [2016-04-17][gym][100947][c][rotate It!]
• Topic: Given a column of numbers, from the first start to take, every other number to take a number, there is an operation is to put the first number in the last, can be unlimited operation, ask the maximum value is how much
• Analysis:
  • Can be seen as a ring, each ring, can start from each ring, ask the maximum value is how much
  • Number is an even number, very simple, is to find the odd position and the sum of the number of pairs of the maximum value
  • Number is an odd number,
  • Analysis can be known that there must be adjacent two numbers in the selected sequence, then only need to enumerate the adjacent two series, calculate the maximum value can be
    • Example of a sample 1 5 3 2 4
    • First 1 4 connected, then 15, one analogy,
    • If you just look at the format of the sample, that is, 1 5 3 2 4, the adjacent two digits are i,j,
    • Assuming I is in an odd position, then the answer becomes the sum of the I-front odd digits and the number of pairs after J.
    • Even the same
• Problems encountered:

  
 

   
  

  
#include<cstdio>
   
  

  
#include<algorithm>
   
  

  
using namespace std;
   
  

  
const int maxn = 1E4 + 10;
   
  

  
long long a[maxn],b[maxn];
   
  

  
int main(){
   
  

  
 int t;
   
  

  
 scanf("%d",&t);
   
  

  
 while(t--){
   
  

  
 int n,tmp;
   
  

  
 long long s[2] = {0};
   
  

  
 scanf("%d",&n);
   
  

  
 if(!(n&1)){
   
  

  
 for(int i = 0; i < n ; ++i){
   
  

  
 scanf("%d",&tmp);
   
  

  
 s[i&1] += tmp;
   
  

  
 }
   
  

  
 printf("%I64d\n",max(s[0],s[1]));
   
  

  
 }else {
   
  

  
 a  [ 0  "    =   b  [ 0  ]    Span class= "pun" >=   0  ;   
   
  

  
 for(int i = 1 ; i <= n; ++i){
   
  

  
 scanf("%d",&tmp);
   
  

  
 if(i & 1){
   
  

  
 a[i] = a[i - 1] + tmp;
   
  

  
 b[i] = b[i - 1];
   
  

  
 }else {
   
  

  
 a[i] = a[i - 1];
   
  

  
 b[i] = b[i - 1] + tmp;
   
  

  
 }
   
  

  
 }
   
  

  
 long long ans = -1E4 * 1E9;
   
  

  
 for(int i = 0 ; i < n ; ++i){
   
  

  
 if(i & 1){
   
  

  
 ans = max(ans , a[i] + b[n] - b[i]); 
   
  

  
 }
   
  

  
 else{
   
  

  
 ans = max(ans , b[i] + a[n] - a[i]); 
   
  

  
 }
   
  

  
 }
   
  

  
 printf("%I64d\n",ans);
   
  

  
 }
   
  

  
 }
   
  

  
 return 0;
   
  

  
}
  
 

From for notes (Wiz)

[2016-04-17] [Gym] [100947] C [Rotate It!]