| # | Time | Username | Problem | Language | Result | Execution time | Memory | 
|---|---|---|---|---|---|---|---|
| 1073786 | horiaboeriu | Kpart (eJOI21_kpart) | C++17 | 704 ms | 856 KiB | 
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <stdio.h>
#include <stdlib.h>
#define MAX 50000
#define MAXN 1000
char f[MAXN + 1];
int v[MAXN + 1], dp[MAX + 1];//dp[x] este cea mai mare pozitie de pe care incepe un subsir cu suma x
int main()
{
    int t1, i1, n, i, sum, x, s, j, nr;
    scanf("%d", &t1);
    for (i1 = 0; i1 < t1; i1++) {
        for (i = 0; i <= MAX; i++) {
            dp[i] = 0;
        }
        scanf("%d", &n);
        sum = 0;
        for (i = 1; i <= n; i++) {
            f[i] = 0;
            scanf("%d", &v[i]);
            sum += v[i];
        }
        sum = sum / 2 + 1;
        //o sa fac dp si o sa iau fiecare secventa din n si ar trebui sa existe un subsir in acea secventa care sa aiba suma cat jumatatea din suma secventei
        for (i = 1; i <= n; i++) {
            for (x = sum - v[i]; x > 0; x--) {
                if (dp[x] > dp[x + v[i]]) {
                    dp[x + v[i]] = dp[x];
                }
            }
            dp[v[i]] = i;//suma v[i] se poate obtine de pe poz i
            s = 0;
            for (j = i; j > 0; j--) {
                s += v[j];
                if (s % 2 > 0 || dp[s / 2] < j) {
                    f[i - j + 1] = 1;
                }
            }
        }
        nr = 0;
        for (i = 1; i <= n; i++) {
            nr += 1 - f[i];
        }
        printf("%d ", nr);
        for (i = 1; i <= n; i++) {
            if (f[i] == 0) {
                printf("%d ", i);
            }
        }
        printf("\n");
    }
    return 0;
}
Compilation message (stderr)
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