UVA586 - Instant Complexity(递归加模拟)

UVA586 - Instant Complexity(递归加模拟)

题目链接

题目大意：给你一段代码，要求你算复杂度。OP代表操作，Loop代表循环，end结束。

解题思路：递归去模拟。具体看代码。

代码:

#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;
}