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| 1 | +using LL = long long; |
| 2 | +class Solution { |
| 3 | +public: |
| 4 | + long long maxTotalValue(vector<int>& nums, int k) { |
| 5 | + int n = nums.size(), K = log2(n) + 1; |
| 6 | + |
| 7 | + int mn[50005][35]; memset(mn, 0x3f, sizeof(mn)); |
| 8 | + int mx[50005][35]; memset(mx, 0xcf, sizeof(mx)); |
| 9 | + for (int i = 0; i < n; i++) mn[i][0] = mx[i][0] = nums[i]; |
| 10 | + for (int k = 1; k <= K; k++) { |
| 11 | + for (int i = 0; i+(1<<k)-1 < n; i++) { |
| 12 | + mn[i][k] = min(mn[i][k-1], mn[i+(1<<(k-1))][k-1]); |
| 13 | + mx[i][k] = max(mx[i][k-1], mx[i+(1<<(k-1))][k-1]); |
| 14 | + } |
| 15 | + } |
| 16 | + |
| 17 | + auto GetDiff = [&](int L, int R){ |
| 18 | + int k = log2(R-L+1); |
| 19 | + return max(mx[L][k], mx[R-(1<<k)+1][k]) - min(mn[L][k], mn[R-(1<<k)+1][k]); |
| 20 | + }; |
| 21 | + |
| 22 | + auto count_ge = [&](int th) { |
| 23 | + int count = 0; |
| 24 | + int j = 0; |
| 25 | + for (int i=0; i<n; i++) { |
| 26 | + while (j<n && GetDiff(i,j)<th) j++; |
| 27 | + count += n-j; |
| 28 | + } |
| 29 | + return count; |
| 30 | + }; |
| 31 | + |
| 32 | + int lo = 0, hi = 2e9; |
| 33 | + while (lo < hi) { |
| 34 | + int mid = hi-(hi-lo)/2; |
| 35 | + if (count_ge(mid) >= k) |
| 36 | + lo = mid; |
| 37 | + else |
| 38 | + hi = mid-1; |
| 39 | + } |
| 40 | + LL th = lo; |
| 41 | + |
| 42 | + LL ret_g = 0; |
| 43 | + LL count_g = 0; |
| 44 | + for (int i=0; i<n; i++) { |
| 45 | + int l = i, r = n-1; |
| 46 | + while (l<r) { |
| 47 | + int mid = l+(r-l)/2; |
| 48 | + if (GetDiff(i,mid)>th) |
| 49 | + r = mid; |
| 50 | + else |
| 51 | + l = mid+1; |
| 52 | + } |
| 53 | + if (GetDiff(i, r) > th) { |
| 54 | + count_g += n-r; |
| 55 | + for (int j=r; j<n; j++) |
| 56 | + ret_g += GetDiff(i, j); |
| 57 | + } |
| 58 | + } |
| 59 | + |
| 60 | + LL ret = ret_g + th * (k-count_g); |
| 61 | + |
| 62 | + return ret; |
| 63 | + |
| 64 | + } |
| 65 | +}; |
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