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91 changes: 91 additions & 0 deletions
Segment_Tree/3671.Sum-of-Beautiful-Subsequences/3671.Sum-of-Beautiful-Subsequences.cpp
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,91 @@ | ||
| class BIT{ | ||
| public: | ||
| int N; | ||
| vector<long long>bitArr; // Note: all arrays are 1-index | ||
| vector<long long>nums; | ||
| long long M = 1e9+7; | ||
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| void init(int N) | ||
| { | ||
| this->N = N; | ||
| bitArr.resize(N+1); | ||
| nums.resize(N+1); | ||
| } | ||
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| // increase nums[i] by delta | ||
| void updateDelta(int i, long long delta) { | ||
| int idx = i; | ||
| while (idx <= N) | ||
| { | ||
| bitArr[idx]+=delta; | ||
| bitArr[idx] %= M; | ||
| idx+=idx&(-idx); | ||
| } | ||
| } | ||
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| // sum of a range nums[1:j] inclusively | ||
| long long queryPreSum(int idx){ | ||
| long long result = 0; | ||
| while (idx){ | ||
| result += bitArr[idx]; | ||
| // result %= M; | ||
| idx-=idx&(-idx); | ||
| } | ||
| return result; | ||
| } | ||
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| // sum of a range nums[i:j] inclusively | ||
| long long sumRange(int i, int j) { | ||
| return queryPreSum(j)-queryPreSum(i-1); | ||
| } | ||
| }; | ||
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| long long MOD = 1e9+7; | ||
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| class Solution { | ||
| public: | ||
| int totalBeauty(vector<int>& nums) { | ||
| int mx = *max_element(begin(nums), end(nums)); | ||
| vector<vector<int>>pos(mx+1); | ||
| for (int i=0; i<nums.size(); i++) { | ||
| for (int j=1; j*j<=nums[i]; j++) { | ||
| if (nums[i]%j==0) { | ||
| pos[j].push_back(i); | ||
| if (j*j!=nums[i]) pos[nums[i]/j].push_back(i); | ||
| } | ||
| } | ||
| } | ||
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| vector<long long>seq(mx+1); | ||
| for (int g=1; g<=mx; g++) { | ||
| map<int,int>mp; | ||
| for (int x: pos[g]) mp[nums[x]] = 1; | ||
| int idx = 0; | ||
| for (auto& [k,v]:mp) v = ++idx; | ||
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| vector<int>arr; | ||
| for (int x: pos[g]) arr.push_back(mp[nums[x]]); | ||
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| BIT bit; | ||
| bit.init(arr.size()); | ||
| for (int x: arr) { | ||
| long long ans = bit.queryPreSum(x-1) + 1; | ||
| seq[g] = (seq[g]+ans)%MOD; | ||
| bit.updateDelta(x, ans); | ||
| } | ||
| } | ||
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| vector<long long>ret(mx+1); | ||
| for (int g=mx; g>=1; g--) { | ||
| ret[g] = seq[g]; | ||
| for (int j=g*2; j<=mx; j+=g) | ||
| ret[g] = (ret[g]-ret[j]+MOD) % MOD; | ||
| } | ||
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| long long ans = 0; | ||
| for (int g=mx; g>=1; g--) { | ||
| ans = (ans + g*ret[g]) % MOD; | ||
| } | ||
| return ans; | ||
| } | ||
| }; |
12 changes: 12 additions & 0 deletions
Segment_Tree/3671.Sum-of-Beautiful-Subsequences/Readme.md
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| @@ -0,0 +1,12 @@ | ||
| ### 3671.Sum-of-Beautiful-Subsequences | ||
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| 此题包含了两个知识点。我们拆开来分析。 | ||
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| 第一个问题,对于任意的正整数g,如何计算在数组里有多少个strictly increasing subsequence并且序列中每个元素都是"g的倍数"。假设我们已经通过预处理,知道数组里g的倍数位于[i1,i2,...,ik]的位置上。注意我们只需要构造严格递增的序列,所以我们需要只知道它们之间的大小关系和位置关系,但是不需要知道绝对大小和绝对位置,故我们可以离散化,转化为类似[1,3,5,2,4]这样的形式,表示数组里有5个数是g的倍数,且它们的相对位置和相对大小都可以用这样的形式表示出来。于是我们就构造出了这样一个问题:里面有多少个严格递增的序列? | ||
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| 这是一个经典问题,我们可以用树状数组来解。我们想象一个长度为5的空数组,需要依照上述次序涂黑。首先我们在填写1时,以它结尾的递增序列只有一个,记作f(1)。然后我们填写3时,以它结尾的递增序列就是`f(3)=f(1)+1`,表示以1结尾的递增序列再附加上3,或者单独的3也可以构成一个符合条件的序列。然后我们填写5时,以它为结尾的递增序列就是`f(5)=f(1)+f(3)+1`。然后我们填写2时,以它结尾的递增序列就是`f(2)=f(1)+1`...由此我们可以看出f(x)就是BIT数组前缀和`f(1)+f(2)+...+f(x-1)`再加1.于是我们就可以求出所有的f(x)并且求和,就得出:数组里有多少个严格递增序列、并且每个元素都是"g的倍数",我们记作p(g). | ||
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| 在有了p(g)的基础上,我们如何进一步求出数组里有多少个严格递增序列、并且所有元素的GCD恰好是g呢?我们记作这样的解是ret(g),其实就是删去在p(g)里去除"2以上g的倍数"的情况,即`ret(g) = p(g)-ret(2g)-ret(3g)-...-ret(kg)`. 所以我们只需要对g按从大到小的顺序求解q:当解到ret(g)时,p(g)和`ret(2g)...ret(kg)`就都是已知的了。初始条件是对于数组里的最大元素mx,`ret(mx) = 1`. | ||
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| 最终返回`g*ret(g)`对于g=1,2,..,mx的之和 | ||
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