Codeforce Round #228 Div2 B

Codeforces Round #229 (Div. 2)
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Submission	Time	Verdict
5965302	02/12/2014 07:09PM	Accepted
5965291	02/12/2014 07:07PM	Wrong answer on test 6
5965246	02/12/2014 07:01PM	Wrong answer on test 4

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Inna is a great piano player and Dima is a modest guitar player. Dima has recently written a song and they want to play it together. Of course, Sereja wants to listen to the song very much.

A song is a sequence of notes. Dima and Inna want to play each note at the same time. At that, they can play the i-th note at volume v (1 ≤ v ≤ a_i; v is an integer) both on the piano and the guitar. They should retain harmony, so the total volume with which the i-th note was played on the guitar and the piano must equal b_i. If Dima and Inna cannot play a note by the described rules, they skip it and Sereja's joy drops by 1. But if Inna and Dima play the i-th note at volumes x_i and y_i (x_i + y_i = b_i) correspondingly, Sereja's joy rises by x_i·y_i.

Sereja has just returned home from the university and his current joy is 0. Help Dima and Inna play the song so as to maximize Sereja's total joy after listening to the whole song!

Input

The first line of the input contains integer n (1 ≤ n ≤ 10⁵) — the number of notes in the song. The second line contains n integers a_i (1 ≤ a_i ≤ 10⁶). The third line contains n integers b_i (1 ≤ b_i ≤ 10⁶).

Output

In a single line print an integer — the maximum possible joy Sereja feels after he listens to a song.

Sample test(s)

Input

3 1 1 2 2 2 3

Output

Input

1 2 5

Output

-1

Note

In the first sample, Dima and Inna play the first two notes at volume 1 (1 + 1 = 2, the condition holds), they should play the last note at volumes 1 and 2. Sereja's total joy equals: 1·1 + 1·1 + 1·2 = 4.

In the second sample, there is no such pair (x, y), that 1 ≤ x, y ≤ 2, x + y = 5, so Dima and Inna skip a note. Sereja's total joy equals -1.

 1 #pragma comment(linker,"/STACK:102400000,102400000")
 2 #include <cstdio>
 3 #include <vector>
 4 #include <cmath>
 5 #include <queue>
 6 #include <set>
 7 #include <cstring>
 8 #include <iostream>
 9 #include <algorithm>
10 using namespace std;
11 #define INF 0x7fffffff 
12 #define mod 1000000007
13 #define ll long long
14 #define maxn 1000005
15 #define pi acos(-1.0)
16 #define dis(a, b) sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y))
17 ll gcd(ll n, ll m){ return m ? gcd(m, n%m) : n; }
18 ll n, m, k, x, y;
19 struct node{int a, b;}p[maxn];
20 bool cmp(node a, node b){ return (a.a < b.a || a.a == b.a&&a.b < b.b); }
21 ll a[maxn], b[maxn], c[maxn];
22 int main(){
23     cin >> n;
24     for (int i = 0; i < n; i++)scanf("%I64d", &a[i]);
25     for (int i = 0; i < n; i++)scanf("%I64d", &b[i]);
26     for (int i = 0; i < n; i++){
27         if (b[i] == 1){ x--; continue; }
28         if (b[i] % 2 == 0 && a[i] >= b[i] / 2)x += b[i] * b[i] / 4;
29         if (b[i] % 2 == 0 && a[i] < b[i] / 2)x--;
30         if (b[i] % 2 == 1 && a[i] > b[i] / 2)x += b[i] / 2 * (b[i] / 2 + 1);
31         if (b[i] % 2 == 1 && a[i] <= b[i] / 2)x--;
32     }
33     cout << x << endl;
34     return 0;
35 }

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posted @ 2014-02-12 23:12 HaibaraAi 阅读(81) 评论(0) 收藏举报

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