Seinfeld杭电3351

题目信息

I’m out of stories. For years I’ve been writing stories, some rather silly, just to make simple problems look difficult and complex problems look easy. But, alas, not for this one.

You’re given a non empty string made in its entirety from opening and closing braces. Your task is to find the minimum number of “operations” needed to make the string stable. The definition for being stable is as follows:
1. An empty string is stable.
2. If S is stable, then {S} is also stable.
3. If S and T are both stable, then ST (the concatenation of the two) is also stable.
All of these strings are stable: {}, {}{}, and {{}{}}; But none of these: }{, {{}{, nor {}{.

The only operation allowed on the string is to replace an opening brace with a closing brace, or visa-versa.

也就是说，把配对的括号忽视掉，剩下不配对的，计算所需要的步骤数，如果是{{或者}}则所需步骤为1，若}{则所需步骤为2，恩，运用栈计算一下就好啦

#include<stdio.h> #include<string.h>#define N 10010char s[N],s1[N];int main(){int lens,i,j,count=0,sum;while(gets(s)){count++;sum=0;if(s[0]=='-')break;lens=strlen(s);s1[0]='!';for(i=0,j=0;i<lens;++i){if(s[i]=='{')s1[++j]=s[i];else if(s[i]=='}'){if(s1[j]=='{')j--;else s1[++j]=s[i];}}if(j==0)printf("%d. 0\n",count);else{for(i=1;i<=j;i+=2){if(s1[i]==s1[i+1])sum+=1;else sum+=2;}printf("%d. %d\n",count,sum);}}return 0;}