9456. Clumsy CowsConstraints
Time Limit: 1 secs, Memory Limit: 256 MB
Description
Bessie the cow is trying to type a balanced string of parentheses into her new laptop, but she is sufficiently clumsy (due to her large hooves) that she keeps mis-typing characters. Please help her by computing the minimum number of characters in the string that one must reverse (e.g., changing a left parenthesis to a right parenthesis, or vice versa) so that the string would become balanced.
There are several ways to define what it means for a string of parentheses to be "balanced". Perhaps the simplest definition is that there must be the same total number of (‘s and )’s, and for any prefix of the string, there must be at least as many (‘s as )’s.
For example, the following strings are all balanced:
()
(())
()(()())
while these are not:
)(
())(
((())))
Input
* Line 1: A string of parentheses of even length at most 100,000 characters.
Output
Line 1: A single integer giving the minimum number of parentheses that must be toggled to convert the string into a balanced string.
Sample Input())(Sample Output2Hint
The last parenthesis must be toggled, and so must one of the two middle right parentheses.
Problem Source
USACO 2012.11/2013年每周一赛第十三场
非常有趣的一道题目：0s：
#include <stdio.h>#include <string.h>//这道题我的思路就是，用right记录没有匹配到的')'，用left动态记录匹配值，是'('加一，是')'减一，然后当left小于零的时候right就要加一（说明出现了一个单独的')'），然后将left清零//注意输入的括号数目确保是偶数char a[100005];int main() {int right = 0, left = 0;gets(a);for (int i = 0; i < (int)strlen(a); i++) {if (a[i] == '(') {left++;} else {left–;}if (left < 0) {right++;left = 0;}}if (right == 0) {if (left == 0) {printf("0\n");//如果两者为零说明全部匹配} else if (left > 0) {printf("%d\n", left / 2);//这种情况说明只剩下'('，只需要把其中一半的'('调转即可}} else {if (left == 0) {printf("%d\n", right / 2);//跟上面的情况差不多} else if (left > 0) {if (left == right) {if (left % 2 == 0) {//这种情况就是这种'))))(((('，，把其中一种括号全部调转即可printf("%d\n", left);} else {printf("%d\n", left + 1);//这种情况是')))))((((('这种情况除了把其中一种括号全部调转外，还要额外调转一个另外一种的括号}} else {if (left % 2 == 0) {//这种情况是'))))))(((('自家括号跟自家匹配，各调转一半printf("%d\n", left / 2 + right / 2);} else {printf("%d\n", (left – 1) / 2 + (right – 1) / 2 + 2);//')))))((('，自家括号跟自家匹配，然后各拿出一个匹配（就是中间那一对）}}}}return 0;}
赚钱之道很多，但是找不到赚钱的种子，便成不了事业家。
