codeforces_667B. Coat of Anticubism

最新推荐文章于 2019-08-07 10:44:49 发布

原创最新推荐文章于 2019-08-07 10:44:49 发布 · 483 阅读

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本博客探讨了一个几何构造问题，即在已知杆长的情况下，通过增加最短杆长来确保可以形成具有非零面积的凸多边形。文章详细介绍了输入数据格式、问题求解逻辑以及输出结果的计算过程。

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As some of you know, cubism is a trend in art, where the problem of constructing volumetrical shape on a plane with a combination of three-dimensional geometric shapes comes to the fore.

A famous sculptor Cicasso, whose self-portrait you can contemplate, hates cubism. He is more impressed by the idea to transmit two-dimensional objects through three-dimensional objects by using his magnificent sculptures. And his new project is connected with this. Cicasso wants to make a coat for the haters of anticubism. To do this, he wants to create a sculpture depicting a well-known geometric primitive — convex polygon.

Cicasso prepared for this a few blanks, which are rods with integer lengths, and now he wants to bring them together. The i-th rod is a segment of length li.

The sculptor plans to make a convex polygon with a nonzero area, using all rods he has as its sides. Each rod should be used as a side to its full length. It is forbidden to cut, break or bend rods. However, two sides may form a straight angle $\text{[math]}$ .

Cicasso knows that it is impossible to make a convex polygon with a nonzero area out of the rods with the lengths which he had chosen. Cicasso does not want to leave the unused rods, so the sculptor decides to make another rod-blank with an integer length so that his problem is solvable. Of course, he wants to make it as short as possible, because the materials are expensive, and it is improper deed to spend money for nothing.

Help sculptor!

Input

The first line contains an integer n (3 ≤ n ≤ 105) — a number of rod-blanks.

The second line contains n integers li (1 ≤ li ≤ 109) — lengths of rods, which Cicasso already has. It is guaranteed that it is impossible to make a polygon with n vertices and nonzero area using the rods Cicasso already has.

Output

Print the only integer z — the minimum length of the rod, so that after adding it it can be possible to construct convex polygon with(n + 1) vertices and nonzero area from all of the rods.

Examples

input
3
1 2 1

output
1

input
5
20 4 3 2 1

output
11

Note

In the first example triangle with sides {1 + 1 = 2, 2, 1} can be formed from a set of lengths {1, 1, 1, 2}.

In the second example you can make a triangle with lengths {20, 11, 4 + 3 + 2 + 1 = 10}.

找几何规律，类似于三角形存在条件的多边形存在条件，也就是最大边长度小于其他各边长度之和。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <stack>
#include <bitset>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <algorithm>
#define Si(a) scanf("%d",&a)
#define Sl(a) scanf("%lld",&a)
#define Sd(a) scanf("%lf",&a)
#define Ss(a) scanf("%s",a)
#define Pi(a) printf("%d\n",(a))
#define Pl(a) printf("%lld\n",(a))
#define Pd(a) printf("%lf\n",(a))
#define Ps(a) printf("%s\n",(a))
#define W(a) while(a--)
#define mem(a,b) memset(a,(b),sizeof(a))
#define FOP freopen("data.txt","r",stdin)
#define inf 0x3f3f3f3f
#define maxn 1000010
#define mod 1000000007
#define PI acos(-1.0)
#define LL long long
using namespace std;

int a[maxn];
int main()
{
    int n,max=0,maxi=0;
    Si(n);
    for(int i=1;i<=n;i++)
    {
        Si(a[i]);
        if(a[i]>max)
        {
            max=a[i];
            maxi=i;
        }
    }
    int sum=0;
    for(int i=1;i<=n;i++)
    {
        if(maxi==i)continue;
        sum+=a[i];
    }
    int ans=0;
    if(sum<=max)ans=max-sum+1;
    Pi(ans);
    return 0;
}