Given an array A
of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K
.
Example 1:
Input: A = [4,5,0,-2,-3,1], K = 5 Output: 7 Explanation: There are 7 subarrays with a sum divisible by K = 5: [4, 5, 0, -2, -3, 1], [5], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3]
Note:
1 <= A.length <= 30000
-10000 <= A[i] <= 10000
2 <= K <= 10000
用了dp,如果某一个数能被K整除,那么以该位结尾的这样的连续子数组的个数就是前一位dp值+1(新的1是它本身)。
如果不能被K整除,那么就让j从i开始往前遍历,碰到第一个从j到i能被K整除的子数组,那么该位上面的长度就是第j位的dp值加上1,新的1指的是从j到i的子数组。
class Solution { public: int subarraysDivByK(vector<int>& A, int K) { vector<int> dp(A.size(),0); vector<int> Bsum(A.size()+1, 0); for(int i=0; i<A.size(); i++){ Bsum[i+1] = Bsum[i] + A[i]; } int ret = 0; for(int i=1; i<Bsum.size(); i++){ if(A[i-1] % K == 0){ if(i-1 == 0) dp[i-1] = 1; else dp[i-1] = dp[i-2] + 1; continue; } int newcnt = 0; for(int j=i-1; j>=0; j--){ //if(Bsum[i] - Bsum[j] == 0) continue; if( (Bsum[i] - Bsum[j]) % K == 0) { //cout << i << " " << j << endl; newcnt += dp[j-1] + 1; break; } } dp[i-1] = newcnt; } for(auto x : dp) ret += x; return ret; } };
原文地址:https://www.cnblogs.com/ethanhong/p/10262354.html
时间: 2024-11-11 11:20:42