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;
}
时间: 2024-10-07 06:47:13