66 lines
1.6 KiB
C
66 lines
1.6 KiB
C
#include <stdio.h>
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#include <stdlib.h>
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#include <string.h>
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// 计算最长合法括号子串长度及数量（数量指达到该最长长度的子串个数）
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// 使用经典 DP：dp[i] 表示以 i 结尾的最长合法括号长度。
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// 当 maxLen 为 0 时，按题意输出 "0 1"。
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int main(void)
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{
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const int MAXN = 1000005; // 题面最大长度约 1e6
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char *s = (char *)malloc(MAXN);
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if (!s) return 0;
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if (scanf("%1000000s", s) != 1) {
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free(s);
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return 0;
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}
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int n = (int)strlen(s);
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if (n == 0) {
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printf("0 1\n");
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free(s);
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return 0;
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}
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int *dp = (int *)calloc(n, sizeof(int));
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if (!dp) {
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free(s);
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return 0;
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}
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int maxLen = 0;
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int count = 1; // 若最终 maxLen 为 0，答案应为 0 1
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for (int i = 1; i < n; ++i) {
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if (s[i] == ')') {
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if (s[i - 1] == '(') {
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dp[i] = 2 + (i >= 2 ? dp[i - 2] : 0);
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} else {
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int prev = i - dp[i - 1] - 1;
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if (prev >= 0 && s[prev] == '(') {
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dp[i] = dp[i - 1] + 2 + (prev >= 1 ? dp[prev - 1] : 0);
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}
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}
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if (dp[i] > 0) {
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if (dp[i] > maxLen) {
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maxLen = dp[i];
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count = 1;
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} else if (dp[i] == maxLen) {
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count++;
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}
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}
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}
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}
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if (maxLen == 0) {
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printf("0 1\n");
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} else {
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printf("%d %d\n", maxLen, count);
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}
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free(dp);
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free(s);
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return 0;
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} |