Codeforces Gym 100889 B Backward and Forward

最新推荐文章于 2020-07-10 18:55:34 发布

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最新推荐文章于 2020-07-10 18:55:34 发布

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分类专栏： Codeforces 文章标签： codeforces

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本文介绍了一种通过合并相邻元素的方法，使数组成为回文数组所需的最少步骤数。具体包括读取数组、比较首尾元素并调整的过程。

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B. Backward and Forward

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output

An array is called a ‘Mirror’ if it reads the same backward and forward. For example, [23, 15, 23] is a ‘Mirror’ but [2, 0, 1] is not.

You are given an array A of size N. Your task is to make a given array a ‘Mirror’. The only allowed operation is that you can merge two adjacent elements in the array and replace them with their sum.

Find minimum number of operations required to make given array a ‘Mirror’ such that it reads the same backward and forward.

Input

First line contains T(1 ≤ T ≤ 20), the number of test cases. Each test case consist of 2 lines. First line contains number N(1 ≤ N ≤ 105), size of the array. Next line contains N space separated integers Ai(0 ≤ Ai ≤ 109) denoting elements in the array A.

Output

For each test case output in one line minimum steps required to make given array a ‘Mirror’.

input

2
3
1 0 1
5
10 20 20 10 40

output

0
3

Sample test case 1:

Given array is already a ‘Mirror’ . So, no need to perform any operation.

Sample test case 2:

One possible sequence of operations.

Step 1 : Merge 2nd and 3rd element to get array [10, 40, 10, 40].

Step 2 : Merge 1st and 2nd element to get array [50, 10, 40].

Step 3 : Merge 2nd and 3rd element to get array [50, 50].

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

__int64 a[100005];

int main()
{
    int t;
    long long n, i, j;
    scanf("%d", &t);
    while(t--){
        long long op = 0;

        scanf("%lld", &n);

        for(i = 0; i < n; i++){
            scanf("%I64d", &a[i]);
        }

        for(i = 0, j = n-1; i <= j; ){
            if(a[i]==a[j]){
                i++;
                j--;
            }
            else if(a[i] > a[j]){ // 需要从尾部开始合并 
                j--;
                a[j]+=a[j+1];
                op++;
            }
            else{   //从头部开始合并 
                i++;
                a[i]+=a[i-1];
                op++;
            }
        }
        printf("%lld\n", op);
    }
    return 0;
}