Blue Bridge Cup: k-fold range

[Blue Bridge Cup 2017 Province B] k times interval

Given a sequence of length N, A1, A2,...AN, if the sum of a continuous subsequence Ai, Ai+1,...Aj is a multiple of K, we call this interval [i,j] K double interval.

Can you find out how many K-fold intervals there are in the sequence?

input format

The first line contains two integers N and K.

Each of the following N lines contains an integer Ai.

output format

Output an integer representing the number of K-fold intervals.

data range

1≤N,K≤100000,

1≤Ai≤100000

Input sample:

5 2

1

2

3

4

5

Sample output:

6

The first method: three layers of violence cycle (timeout)

#include<iostream>
using namespace std;
int main()
{
    int n, g[100010], i, j, k, sum = 0, ans = 0;
    cin >> n >> k;
    for (i = 1; i <= n; i++)
    {
        cin >> g[i];
    }
    for (i = 1; i <= n; i++)
    {
        for (j = i; j <= n; j++)
        {
            sum = 0;
            for (int m = i; m <= j; m++)
            {
                sum += g[m];
            }
            if (sum % k == 0)
                ans++;
        }
    }
    cout << ans << endl;
    return 0;
}

The second method: two layers of loop + prefix and (still timeout)

#include<iostream>
using namespace std;
int main()
{
    int n, g[100010], s[100010]={0}, i, j, k, sum = 0, ans = 0;
    cin >> n >> k;
    for (i = 1; i <= n; i++)
    {
        cin >> g[i];
        s[i] = s[i - 1] + g[i];
    }
    for (i = 1; i <= n; i++)
    {
        for (j = i; j <= n; j++)
        {
            int t = s[j] - s[i - 1];
            if (t % k == 0)
                ans++;
        }
    }
    cout << ans << endl;
    return 0;
}

The third method: one layer of loop + prefix sum (no timeout)

#include<cstdio>
#include<iostream>
using namespace std;
typedef long long ll;
const int N = 100010;
int n, k;
ll s[N], cnt[N];

int main()
{
    scanf("%d %d", &n, &k);
    for (int i = 1; i <= n; i++)
    {
        scanf("%lld", &s[i]);
        s[i] += s[i - 1];
    }
    ll res = 0;
    cnt[0] = 1;
    for (int i = 1; i <= n; i++)
    {
        res += cnt[s[i] % k];
        cnt[s[i] % k]++;
    }
    printf("%lld\n", res);
    return 0;
}