The Contest CodeForces - 813A (思维)

最新推荐文章于 2024-12-01 03:34:23 发布

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最新推荐文章于 2024-12-01 03:34:23 发布

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原文链接：http://www.cnblogs.com/qieqiemin/p/10256416.html

本文详细解析了一项在线编程竞赛的策略，通过合理安排问题解决和提交时机，以最小化总时间消耗。讨论了如何在有限的工作时间内，优化多个问题的解决顺序，以达到最佳竞赛成绩。并提供了一个具体的代码示例，展示了如何根据问题难度和网站可用时间窗口来计算最短完成时间。

Pasha is participating in a contest on one well-known website. This time he wants to win the contest and will do anything to get to the first place!

This contest consists of n problems, and Pasha solves ith problem in a_i time units (his solutions are always correct). At any moment of time he can be thinking about a solution to only one of the problems (that is, he cannot be solving two problems at the same time). The time Pasha spends to send his solutions is negligible. Pasha can send any number of solutions at the same moment.

Unfortunately, there are too many participants, and the website is not always working. Pasha received the information that the website will be working only during m time periods, jth period is represented by its starting moment l_j and ending moment r_j. Of course, Pasha can send his solution only when the website is working. In other words, Pasha can send his solution at some moment T iff there exists a period x such that l_x ≤ T ≤ r_x.

Pasha wants to know his best possible result. We need to tell him the minimal moment of time by which he is able to have solutions to all problems submitted, if he acts optimally, or say that it's impossible no matter how Pasha solves the problems.

Input

The first line contains one integer n (1 ≤ n ≤ 1000) — the number of problems. The second line contains n integers a_i (1 ≤ a_i ≤ 10⁵) — the time Pasha needs to solve ith problem.

The third line contains one integer m (0 ≤ m ≤ 1000) — the number of periods of time when the website is working. Next m lines represent these periods. jth line contains two numbers l_j and r_j (1 ≤ l_j < r_j ≤ 10⁵) — the starting and the ending moment of jth period.

It is guaranteed that the periods are not intersecting and are given in chronological order, so for every j > 1 the condition l_j > r_j - 1 is met.

Output

If Pasha can solve and submit all the problems before the end of the contest, print the minimal moment of time by which he can have all the solutions submitted.

Otherwise print "-1" (without brackets).

Examples

Input

Output

Input

Output

-1

Input

Output

Note

In the first example Pasha can act like this: he solves the second problem in 4 units of time and sends it immediately. Then he spends 3 time units to solve the first problem and sends it 7 time units after the contest starts, because at this moment the website starts working again.

In the second example Pasha invents the solution only after the website stops working for the last time.

In the third example Pasha sends the solution exactly at the end of the first period.

思路：

注意题干中的这句话，The time Pasha spends to send his solutions is negligible. Pasha can send any number of solutions at the same moment.

这样的话，我们就可以把所有的题目全部做完后然后在最优的时间进行统一提交，

那么把所有思考问题的时间进行求前缀和，然后和区间的l和r进行比较即可得出答案。

我的AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define gg(x) getInt(&x)
using namespace std;
typedef long long ll;
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
ll n,m;
ll x;
ll l,r;
int main()
{
    gbtb;
    cin>>n;
    ll sum=0ll;
    repd(i,1,n)
    {
        cin>>x;
        sum+=x;
    }
    cin>>m;
    int flag=0;
    ll ans;
    repd(i,1,m)
    {
        cin>>l>>r;
        if(flag==0&&sum>=l&&sum<=r)
        {
            flag=1;
            ans=sum;
        }else if(flag==0&&l>sum)
        {
            flag=1;
            ans=l;
        }

    }
    if(flag)
    {
        cout<<ans<<endl;
    }else
    {
        cout<<-1<<endl;
    }
    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}