A B D均比较简单。先说C题
Given an integer array arr and an integer k, modify the array by repeating it k times.
For example, if arr = [1, 2] and k = 3 then the modified array will be [1, 2, 1, 2, 1, 2].
Return the maximum sub-array sum in the modified array. Note that the length of the sub-array can be 0 and its sum in that case is 0.
As the answer can be very large, return the answer modulo 10^9 + 7.
Example 1:
Input: arr = [1,2], k = 3 Output: 9
Example 2:
Input: arr = [1,-2,1], k = 5 Output: 2
Example 3:
Input: arr = [-1,-2], k = 7 Output: 0
Constraints:
1 <= arr.length <= 10^51 <= k <= 10^5-10^4 <= arr[i] <= 10^4这里有个猜测，
如果K=1，那么我只需要算这个数组的最大子数组和即可。
如果K=2，那么我也只需要算这个数组的最大子数组和，只不过是多加了长度。
如果K>2，那么不需要再讨论什么子数组了，实际上和K<=2类似。所以K>2以后的数组，如果和大于0，则全都要。
class Solution { public: int kConcatenationMaxSum(vector<int>& arr, int k) { int sum1 = 0,sum2 = 0; int ant = 0; int sum3 = 0; int arrsize = arr.size(); for(int i = 0;i<arrsize;i++){ ant+=arr[i]; sum3+=arr[i]; sum1 = max(ant,sum1); if(ant<=0){ ant=0; } } ant = 0; for(int i = 0;i<arrsize*2;i++){ ant+=arr[i%arrsize]; sum2 = max(ant,sum2); if(ant<=0){ ant=0; } } if(k==1){ return sum1; } cout<<sum3<<" "<<sum1<<" "<<sum2<<endl; if(sum3>0){ for(int i=2;i<k;i++){ sum2 +=sum3; sum2%=1000000007; } } return sum2; } };
原文：https://www.cnblogs.com/yinghualuowu/p/11525205.html