UVA 586 - Instant Complexity(dfs)

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UVA 586 - Instant Complexity

题目链接

题意：给定一些操作，问这些操作一共执行了多少步

思路：按操作dfs下去，然后每次END的时候加上对应的幂次的个数，最后注意输出的几个坑点即可

代码：

#include <cstdio>#include <cstring>int t, ans[15];char c[10], str[105];void dfs(int x, int k) {    int sum = 0;    while (~scanf("%s", c)) {if (c[0] == 'E') {    ans[k] += x * sum;    return;}int num;scanf("%s", str);if (c[0] == 'L') {    if (str[0] == 'n')dfs(x, k + 1);    else {sscanf(str, "%d", &num);dfs(x * num, k);    }}else {    sscanf(str, "%d", &num);    sum += num;}    }}int main() {    int cas = 0;    scanf("%d", &t);    while (t--) {scanf("%s", c);memset(ans, 0, sizeof(ans));dfs(1, 0);printf("Program #%d\n", ++cas);printf("Runtime = ");int bo = 0;int flag = 1;for (int i = 10; i > 0; i--) {    if (!ans[i]) continue;    flag = 0;    if (bo++) printf("+");    if (ans[i] != 1) printf("%d*", ans[i]);    if (i == 1) printf("n");    else printf("n^%d", i);}if (ans[0]) {    flag = 0;    if (bo++) printf("+");    printf("%d", ans[0]);}if (flag) printf("0");printf("\n\n");    }    return 0;}

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