6609 - Minimal Subarray Length

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4620

 

 

You are given an integer sequence of length N and another value X. You have to find a contiguous
subsequence of the given sequence such that the sum is greater or equal to X. And you have to find
that segment with minimal length.
Input
First line of the input file contains T the number of test cases. Each test case starts with a line
containing 2 integers N (1 ≤ N ≤ 500000) and X (−109 ≤ X ≤ 109
). Next line contains N integers
denoting the elements of the sequence. These integers will be between −109
to 109
inclusive.
Output
For each test case output the minimum length of the sub array whose sum is greater or equal to X. If
there is no such array, output ‘-1’.
Sample Input
3
5 4
1 2 1 2 1
6 -2
-5 -6 -7 -8 -9 -10
5 3
-1 1 1 1 -1
Sample Output
3
-1
3

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;

const int N = 510000;
const int INF = 0x3fffffff;
typedef long long LL;
#define met(a,b) (memset(a,b,sizeof(a)))

struct node
{
    LL x;
    int Start;
} sum[N];

LL a[N];

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int n, i, Min=N, X, Start;
        LL x;

        met(a, 0);
        met(sum, 0);

        scanf("%d%lld", &n, &x);

        for(i=1; i<=n; i++)
            scanf("%lld", &a[i]);

        for(i=1; i<=n; i++)
        {
            if(sum[i-1].x<=0 || i==1)
            {
                sum[i].x = a[i];
                sum[i].Start = i;
            }
            else
            {
                sum[i].x = sum[i-1].x + a[i];
                sum[i].Start = sum[i-1].Start;
            }
            if(sum[i].x>=x)
            {
                Min = min(Min, i-sum[i].Start+1);
                X = sum[i].x, Start = sum[i].Start;
                while(X>=0 && Start<=i)
                {
                    X -= a[Start];
                    Start++;
                    if(X >= x)
                    {
                        sum[i].x = X;
                        sum[i].Start = Start;
                        Min = min(Min, i-Start+1);
                    }
                }
            }
        }

        printf("%d\n", Min!=N?Min:-1);
    }

    return 0;
}


/**

300
5 4
1 2 1 2 1
6 -2
-5 -6 -7 -8 -9 -10
5 3
-1 1 1 1 -1
8 6
1 1 1 1 1 2 3 4
6 5
4 -3 4 -1 2 2
6 6
-5 1 2 4 1 3
6 5
4 -3 4 -1 -2 2
6 5
-1 -1 -2 3 -2 5
4 5
3 -2 4 1
8 6
1 1 1 1 1 3 1 2
8 6
1 1 1 1 1 3 2 1
8 6
1 1 1 1 1 3 1 1


**/