Submission #1109933

#	Time	Username	Problem	Language	Result	Execution time	Memory
1109933		parsadox2	Fancy Fence (CEOI20_fancyfence)	C++17	55 / 100	1085 ms	4688 KiB

fancyfence

#include<bits/stdc++.h>
using namespace std;
#define int long long

const int MOD = 1e9+7;



void solve() {
    int n;
    cin >> n;
    pair<int,int> nums[n];
    int mx = 0;
    bool b = 1, b2 = 1;
    for (int i = 0;i < n;i++) {
        cin >> nums[i].first;
        mx = max(mx, nums[i].first);
    }
    for (int i = 1;i < n;i++) {
        if (nums[i].first != nums[i-1].first) {
            b = false;
        }
        if (nums[i].first < nums[i-1].first) {
            b2 = false;
        }
    }
    for (int i = 0;i < n;i++) {
        cin >> nums[i].second;
    }
    int ans = 0;
    if (b) {
        int len = 0;
        for (int i = 0;i < n;i++) {
            len += nums[i].second;
        }
        len%= MOD;
        ans = (((mx)*(mx+1)/2)%MOD)*(((len)*(len+1)/2)%MOD);
        ans %= MOD;
        cout << ans;
        return;
    }
    if (b2) {
        int suf[n];
        suf[n-1] = nums[n-1].second;
        for (int i = n-2;i >= 0;i--) {
            suf[i] = (suf[i+1] + nums[i].second)%MOD;
        }
        int last = 0;
        int ans = 0;
        for (int i = 0;i < n;i++) {
            int x = (  (((nums[i].first+1)* (nums[i].first)/2)%MOD) - (((last+1)* (last)/2)%MOD)  + MOD )%MOD;
            int y = (((suf[i]+1)*(suf[i])/2)%MOD);
            ans = (ans + (x*y))%MOD;
            last = nums[i].first;
        }
        cout << ans;
        return;
    }
    for (int h = 1;h <= mx;h++) {
        int len = 0;
        for (int i = 0;i < n;i++) {
            if (nums[i].first < h) {
                len %= MOD;
                ans = (ans + (h * (((len)*(len+1)/2)%MOD)))%MOD;
                len = 0;
            } else {
                len += nums[i].second;
            }
        }
        len %= MOD;
        ans = (ans + ((h) * (((len)*(len+1)/2)%MOD)))%MOD;
        len = 0;
    }
    cout << ans;
}


int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int tc = 1;
    //cin >> tc;
    while (tc--) {
        solve();
    }
}

• Subtask 1: 0 / 0
• Subtask 2: 12 / 12
• Subtask 3: 13 / 13
• Subtask 4: 15 / 15
• Subtask 5: 15 / 15
• Subtask 6: 0 / 18
• Subtask 7: 0 / 27