Uva586-instant complexity (recursive and simulated)

Uva586-instant complexity (recursive and simulated)

Question Link

A piece of code requires you to calculate complexity. OP indicates the operation, loop indicates the loop, and end indicates the end.

Solution: recursive simulation. For more information, see the code.

Code:

#include <cstdio>#include <cstring>const int N = 15;char s1[N], s2[N];typedef long long ll;ll nv[N];void solve (ll * v) {    while (1) {        scanf ("%s", s1);        int num;        if (s1[0] == ‘O‘) {            scanf ("%s", s2);            if (s2[0] != ‘n‘) {                sscanf (s2, "%d", &num);                for (int i = 0; i < N; i++)                    if (v[i])                        nv[i] += v[i] * num;             } else {                for (int i = 0; i < N - 1; i++)                        if (v[i])                        nv[i + 1] += v[i];            }        } else if (s1[0] == ‘L‘) {            scanf ("%s", s2);            ll tmp[N];            memset (tmp, 0, sizeof (tmp));            if (s2[0] != ‘n‘) {                sscanf (s2, "%d", &num);                for (int i = 0; i < N; i++)                    tmp[i] = num * v[i];            } else {                for (int i = 0; i < N - 1; i++)                    tmp[i + 1] = v[i];            }                solve(tmp);        } else if (s1[0] == ‘E‘)            break;    }    }int main () {    int T, num;    scanf ("%d", &T);    for (int cas = 1; cas <= T; cas++) {        memset (nv, 0, sizeof (nv));        ll v[N];        memset (v, 0, sizeof (v));        v[0] = 1;        solve(v);        //PRINTF        printf ("Program #%d\n", cas);        printf ("Runtime = ");        bool flag = 0;        for (int j = N - 1; j >= 0; j--) {            if (nv[j]) {                if (flag)                    printf ("+");                if (nv[j] == 1) {                    if (j == 1)                        printf ("n", j);                    else if (j > 1)                        printf ("n^%d", j);                    else                        printf ("%lld", nv[j]);                } else {                    if (j == 1)                        printf ("%lld*n", nv[j]);                    else if (j > 1)                        printf ("%lld*n^%d", nv[j], j);                    else                        printf ("%lld", nv[j]);                }                flag = 1;            }            }        if (!flag)            printf ("0");        printf ("\n\n");    }    return 0;}

Uva586-instant complexity (recursive and simulated)