Greedy Algorithm-maximum sum of at least K consecutive intervals

For an integer N, it indicates the number of integers in the interval. Finding at least K consecutive numbers is the largest sum of the segments in the interval.

The range of N is 1 ~ 10 ^ 6. K (-1000 ~ 1000 );

Analysis: This question looks a bit like a tree array, but if you want to find the sum of at least K consecutive numbers in the interval is the largest, it may be a bit difficult. At most, it is the sum of a certain range.

However, we can follow this idea: sum (N)-sum (I )?.

If we think like this, we need to judge n-I> = K; then another way: Find the minimum value before I and find the maximum value after I + K. In this way, sum (N)-sum (I) must be the largest in the I-th loop.

For every such loop, I only need to define a variable to receive a maximum value (sum (N)-sum (I) and compare it with the next one.

Clear thinking,CodeIt should be easy to write.

 # include 
  
    # include 
   
     using namespace STD; int N, K; int A [1000005]; int INF = 0x7fffffff; void maxsum () {int L = inf; // The maximum value, can pass the value to L (less than its value ). Int r = inf + 1; // The minimum value. It can pass the value to R (greater than its value ). For (INT I = K; I <= N; I ++) {L = L 
     A [I]-l? R: A [I]-l; // check the maximum value after I.} Cout 
     
       A [I]; A [I] + = A [I-1];} maxsum ();} return 0;}