Submission #650267

#TimeUsernameProblemLanguageResultExecution timeMemory
650267alvinpiterFancy Fence (CEOI20_fancyfence)C++17
100 / 100
102 ms3296 KiB
/* Assume we already know the number of rectangles on the first (i - 1) fence. Define f(i) as (i * i + i)/2; If we are to add the i-th fence, then the new rectangles can be categorized into 3: * It lies completely on the i-th fence. There are f(h[i]) * f(w[i]) such rectangles. * The right end ends on the i-th fence while the left end lies on taller fence to the left of it. We don't want to consider the taller fence that comes before a fence lower than the i-th fence. Let's say the first lower fence to the left of the i-th fence is the j-th fence. There are (sumWidth(j + 1, i) * w[i]) * f(h[i]) such rectangles. * The right end ends on the i-th fence while the left end lies on lower fences. Assume we keep the lower fences on a stack: h_i1 < h_i2 < h_i3 ... We can create rectangles whose height is at most h_i3 using the (i2 + 1)-th fence upto the i3-th fence. There are (sumWidth(i2 + 1, i3) * w[i]) * f(h_i3) such rectangles. We can do the same calculation for the (i1 + 1)-th fence upto the i2-th fence, and so on. */ #include<bits/stdc++.h> using namespace std; #define LL long long int #define MOD 1000000007 #define MAXN 100000 int n; LL h[MAXN + 3], w[MAXN + 3], sumWidth[MAXN + 3], sumFhw = 0, ans = 0; stack<int> st; LL f(LL k) { LL k1 = k + 1; return k%2 == 0 ? ((k/2)%MOD * (k1)%MOD)%MOD : (k%MOD * (k1/2)%MOD)%MOD; } int main() { cin >> n; for (int i = 1; i <= n; i++) { cin >> h[i]; } sumWidth[0] = 0; for (int i = 1; i <= n; i++) { cin >> w[i]; sumWidth[i] = sumWidth[i - 1] + w[i]; } for (int i = 1; i <= n; i++) { LL rectanglesOnTheCurrentFence = (f(h[i]) * f(w[i]))%MOD; ans = (ans + rectanglesOnTheCurrentFence)%MOD; while (!st.empty() && h[st.top()] >= h[i]) { int top = st.top(); st.pop(); LL totalWidth = sumWidth[top] - (st.empty() ? 0 : sumWidth[st.top()]); LL fhw = (f(h[top]) * (totalWidth%MOD))%MOD; sumFhw = (sumFhw - fhw + MOD)%MOD; } int firstLowerIdx = (st.empty() ? 0 : st.top()); LL rectanglesOnTheCurrentFenceAndTallerNeighbors = (((sumWidth[i - 1] - sumWidth[firstLowerIdx])%MOD * w[i])%MOD * f(h[i]))%MOD; ans = (ans + rectanglesOnTheCurrentFenceAndTallerNeighbors)%MOD; LL rectanglesOnTheCurrentFenceAndLowerNeighbors = (sumFhw * w[i])%MOD; ans = (ans + rectanglesOnTheCurrentFenceAndLowerNeighbors)%MOD; LL totalWidth = sumWidth[i] - (st.empty() ? 0 : sumWidth[st.top()]); sumFhw = (sumFhw + (f(h[i]) * (totalWidth%MOD))%MOD)%MOD; st.push(i); } cout << ans << endl; }
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