Down the Pyramid

最新推荐文章于 2022-05-02 16:35:13 发布

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最新推荐文章于 2022-05-02 16:35:13 发布

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原文链接：http://www.cnblogs.com/tingtin/p/9337842.html

本文探讨了一种逆向构造数塔的问题，即已知数塔顶部序列，求解底部序列的可能性。通过数学分析确定了可能解的数量，并提供了一个有效的算法实现。

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Do you like number pyramids? Given a number sequence that represents the base, you are usually supposed to build the rest of the "pyramid" bottom-up: For each pair of adjacent numbers, you would compute their sum and write it down above them. For example, given the base sequence

However, I am not interested in completing the pyramid – instead, I would much rather go underground. Thus, for a sequence of

Input Format

The input consists of:

one line with the integer $\le n \le 10^6)(1≤n≤106), the length of the base sequence.$
one line with n integers $a_1, \cdots , a_na1,⋯,an (0 \le ai \le 10^8(0≤ai≤108 for each i)i), forming the base sequence.$

Output Format

Output a single integer, the number of non-negative integer sequences that would have the input sequence as the next level in a number pyramid.

样例输入1

6
12 5 7 7 8 4

样例输出1

样例输入2

3
10 1000 100

样例输出2

题目来源

German Collegiate Programming Contest 2018

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <utility>
#include <vector>
#include <map>
#include <queue>
#include <stack>
#include <cstdlib>
#include <cmath>
typedef long long ll;
#define lowbit(x) (x&(-x))
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
using namespace std;
#define pi acos(-1)
int n;
const int N=1e6+9;
const ll  inf=1e14+4;
ll a[N];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%lld",&a[i]);
    }
    ll  l=0,r=0;
    ll maxx=inf,minn=0;//一定要>=0
    for(int i=1;i<=n;i++){
        if(i%2==1){
            l=a[i]-r;
            maxx=min(maxx,l);
        }
        if(i%2==0){
            r=a[i]-l;
            minn=max(minn,r*(-1));
        }
    }
    //解的范围[minn,maxx]
    /*
    如 12 5 7 11 
    x 12-x -7+x 14-x  x-3
    令x==0 0 12 -7 14 -3
    [7,12]
    */
    
    if(minn>maxx){
        printf("0\n");
    }
    else{
        printf("%lld\n",maxx-minn+1);
    }
    return 0;
}