Codeforces Round #229 (Div. 2) B. Inna, Dima and Song

B. Inna, Dima and Song

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 byx_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

31 1 22 2 3

output

4

input

125

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.

题目是很简单的。。。但是我又给long long 忘了，到了最后还剩半小时的时候才想起来

每次判断是否存在xi yi满足和为bi，若满足，开心度会上升xi*yi（取最大者）

很简单的一个一元二次方程求最大值问题

代码如下：

#include <algorithm>#include <iostream>#include <string>#include <vector>#define maxn 100010using namespace std;int a[maxn],b[maxn];int main(void){    int n;    while(cin >> n){     for(int i=0; i<n; ++i)      cin >> a[i];     for(int i=0; i<n; ++i)      cin >> b[i];           long long joy = 0;          for(int i=0; i<n; ++i)      if(2*a[i] >= b[i] && b[i]>=2){       long long v = b[i]/2;       long long t = b[i]-v;       joy = joy + (v*t);      }      else --joy;     cout << joy << endl;    }    return 0;}