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722e789
Create 3447.Assign-Elements-to-Groups-with-Constraints.cpp
wisdompeak dd363d6
Update Readme.md
wisdompeak c71d7fc
Create Readme.md
wisdompeak 2318d0e
Create 3448.Count-Substrings-Divisible-By-Last-Digit.cpp
wisdompeak 2916a52
Update Readme.md
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43 changes: 43 additions & 0 deletions
...ount-Substrings-Divisible-By-Last-Digit/3448.Count-Substrings-Divisible-By-Last-Digit.cpp
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,43 @@ | ||
| using LL = long long; | ||
| class Solution { | ||
| int n; | ||
| public: | ||
| long long countSubstrings(string s) | ||
| { | ||
| n = s.size(); | ||
| vector<int>nums; | ||
| for (auto ch: s) | ||
| nums.push_back(ch-'0'); | ||
| nums.insert(nums.begin(), 0); | ||
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| LL ret = 0; | ||
| for (int k=1; k<=9; k++) | ||
| ret += helper(nums, k); | ||
| return ret; | ||
| } | ||
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| LL helper(vector<int>&nums, int k) | ||
| { | ||
| vector<LL>count(k, 0); | ||
| vector<LL>count2(k,0); | ||
| LL ret = 0; | ||
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| int r = 0; | ||
| count[0] = 1; | ||
| for (int i=1; i<=n; i++) | ||
| { | ||
| for (int d=0; d<k; d++) | ||
| count2[d] = 0; | ||
| for (int d=0; d<k; d++) | ||
| count2[(d*10)%k]+=count[d]; | ||
| count = count2; | ||
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| r = (r*10+nums[i])%k; | ||
| if (nums[i]==k) | ||
| ret += count[r]; | ||
| count[r]+=1; | ||
| } | ||
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| return ret; | ||
| } | ||
| }; |
29 changes: 29 additions & 0 deletions
...n-Elements-to-Groups-with-Constraints/3447.Assign-Elements-to-Groups-with-Constraints.cpp
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,29 @@ | ||
| class Solution { | ||
| public: | ||
| vector<int> assignElements(vector<int>& groups, vector<int>& elements) | ||
| { | ||
| int n = *max_element(groups.begin(), groups.end()); | ||
| vector<int>arr(n+1, -1); | ||
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| for (int j=0; j<elements.size(); j++) | ||
| { | ||
| int x0 = elements[j]; | ||
| if (x0>n) continue; | ||
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| if (arr[x0]!=-1) continue; | ||
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| int x = x0; | ||
| while (x<=n) | ||
| { | ||
| if (arr[x]==-1) | ||
| arr[x] = j; | ||
| x+=x0; | ||
| } | ||
| } | ||
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| vector<int>rets; | ||
| for (int g: groups) | ||
| rets.push_back(arr[g]); | ||
| return rets; | ||
| } | ||
| }; |
7 changes: 7 additions & 0 deletions
Others/3447.Assign-Elements-to-Groups-with-Constraints/Readme.md
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,7 @@ | ||
| ### 3447.Assign-Elements-to-Groups-with-Constraints | ||
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| 突破点在于groups里的元素的数值不超过1e5.在这个范围是,如果枚举所有1的倍数,然后枚举所有2的倍数,然后枚举所有3的倍数,直至枚举n的倍数,那么总共的时间复杂度是`n+n/2+n/3+...n/n = n*(1+1/2+1/3+...1/n)`.这个级数虽然不收敛,但是它是趋近于nlog(n)的。所以本题可以用暴力枚举。 | ||
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| 所以本题的算法很简单。我们开一个长度为1e5的数组assign,来记录每个自然数最早能被哪个element所assign。我们依次考察element里的每个元素,比如说elements[j]=x,然后枚举x的所有倍数(直至1e5),比如说kx,那样就有`assign[kx] = j`,当然根据题意,我们对于每个assign我们只更新一次。 | ||
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| 最后根据groups的数值,从assgin里把答案拷贝过去即可。 |
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