Accordion
https://codeforces.com/contest/1101/problem/B
題解:
從左往右找到?[: 記:下標為l
從右往左找到?:]?記:下標為r
查找【l+1,r-1】區間上的?|?數量cnt
答案就是cnt+4
如果不能滿足l<r或者找不到?[: 和?:]任意一個那么不存在?
/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,q; int ans,cnt,flag,temp; int a[N]; string str; stack<char>st1; stack<char>st2; int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//scanf("%d",&n);//scanf("%d",&t);//while(t--){}cin>>str;int len=str.size();flag=0;int l=0,r=len-1;for(int i=0;i<len;i++){if(str[i]=='['&&st1.size()==0){st1.push('[');}if(str[i]==':'&&st1.size()==1){l=i;st1.push(':');break;}}for(int i=len;i>0;i--){if(str[i]==']'&&st2.size()==0){st2.push(']');}if(str[i]==':'&&st2.size()==1){r=i;st2.push(':');break;}}if(st1.size()!=2||st2.size()!=2||l>=r||l==0||r==len-1){cout<<-1<<endl;}else{for(int i=l+1;i<r;i++){if(str[i]=='|'){cnt++;}}cout<<cnt+4<<endl;}//cout << "Hello world!" << endl;return 0; }?
?
總結
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