Codeforces 260B - Ancient Prophesy

260B - Ancient Prophesy

思路：字符串处理，把符合条件的答案放进map里，用string类中的substr()函数会简单一些，map中的值可以边加边记录答案，可以省略迭代器访问部分。

代码：

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
map<string,int>mp;
string s;
int d[12]={31,28,31,30,31,30,31,31,30,31,30,31};
bool isOK(string s)
{if(s[2]!='-'||s[5]!='-')return false;if(s[0]=='-'||s[1]=='-'||s[3]=='-'||s[4]=='-')return false;int a=(s[0]-'0')*10+s[1]-'0';int b=(s[3]-'0')*10+s[4]-'0';if(b<=0||b>12)return false;if(a<=0||a>d[b-1])return false;return true;
}
int main()
{cin>>s;int cnt=0;string ans;for(int i=6;i<s.size()-3;i++){if(s[i]=='2'&&s[i+1]=='0'&&s[i+2]=='1'&&'3'<=s[i+3]&&s[i+3]<='5'){string s1=s.substr(i-6,6);if(isOK(s1)){string s2=s.substr(i-6,10);mp[s2]++;if(mp[s2]>cnt){cnt=mp[s2];ans=s2;}}}}cout<<ans<<endl;return 0;
}

转载于:https://www.cnblogs.com/widsom/p/7205556.html

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