Set weights for each subarray and get the response of each subarray. Put the weights in a cell array. wts1 = ones(nrows*n1,1); wts2 = 1.5*ones(nrows*n2,1); wts3 = 3*ones(nrows*n3,1); resp = partarray(fc,[30;0],c,{wts1,wts2,wts3}) ...
此题的子问题可以定为 maxSubArray(vector<int> &nums, int i); 就是在子序列nums[0:i] 中的最大序列。 classSolution {public:intmaxSubArray(vector<int>&nums) {intn =nums.size();if(n ==1)returnnums[0];//if (n == 2) return max(nums[0],nums[1]);vector<int>dp(n); dp[0] = ...
steer_ang_subarrays = [5 15 30;0 0 0]; sv_subarray = phased.SteeringVector('SensorArray',subarray,...'PropagationSpeed',c); wc = sv_subarray(fc,steer_ang_subarrays); array.SubarraySteering ='Custom'; pattern(array,fc,-90:90,0,'CoordinateSystem','Rectangular',...'Type','powerdb'...
#include <bits/stdc++.h> using namespace std; class Solution { public: int numSubarrayBoundedMax(vector<int>& A, int L, int R) { int ret = 0; int dp = 0; int prev = -1; for(int i = 0; i < A.size(); i++){ if(A[i] < L && i > 0){ ret += dp; } if(A[i...
A in the input array, n is the length of the array & s in the given sum. Initialize vector b. (for storing indexes of subarray) Initialize a variable cur_sum=0 for i=0:n-1 set cur_sum=A[i] for j=i+1:n-1 add A[j] to cur_sum if(cur_sum==s) i is the starting index...
classSolution {public:boolcheckSubarraySum(vector<int>& nums,intk) {intn = nums.size(), sum =0; unordered_map<int,int> m{{0,-1}};for(inti =0; i < n; ++i) { sum+=nums[i];intt = (k ==0) ? sum : (sum %k);if(m.count(t)) {if(i - m[t] >1)returntrue; ...
less than 5.06% of C++ online submissions for Maximum Sum of 3 Non-Overlapping Subarrays.classSolution{public:vector<int>maxSumOfThreeSubarrays(vector<int>&nums,intk){constintN=nums.size();constintM=N-k+1;vector<int>sums(M);sums[0]=accumulate(begin(nums),begin(nums)+k,0);for(inti=...
vector<PII>vv[MAX];//gcd值和对应的位置 inta[MAX]; intpos[MAX]; intans[MAX]; intc[MAX]; voidupdate(intx,intval) { while(x<MAX) { c[x]+=val; x+=x&-x; } } intsum(intx) { intres=0; while(x>0) { res+=c[x];
#include<vector> using namespace std; int a[100010]; int sum[100010]; map<int,int>mp; int main() { int t; cin>>t; while(t--) { mp.clear(); int n; cin>>n; string s; cin>>s; for(int i=0;i<n;i++) a[i+1]=s[i]-'0' - 1; ...
classSolution {public:intcnt(vector<int>& A,vector<int>& B,inti) {intc =0, last =0;for(intj =0; j < B.size(); j++) {if(c > last) last = c;if(i + j <0)continue;if(i + j >= A.size()) {break; }if(A[i + j] == B[j]) c++;if(c > last) last = c;if(...