题目描述
We have N voting papers. The i-th vote (1≤i≤N) has the string Si written on it.
Print all strings that are written on the most number of votes, in lexicographical order.
Constraints
·1≤N≤2×105
·Si (1≤i≤N) are strings consisting of lowercase English letters.
·The length of Si (1≤i≤N) is between 1 and 10 (inclusive).
输入
Input is given from Standard Input in the following format: N S1 : SN
输出
Print all strings in question in lexicographical order.
样例输入
【样例1】 7 beat vet beet bed vet bet beet 【样例2】 8 buffalo buffalo buffalo buffalo buffalo buffalo buffalo buffalo 【样例3】 7 bass bass kick kick bass kick kick 【样例4】 4 ushi tapu nichia kun
样例输出
【样例1】 beet vet 【样例2】 buffalo 【样例3】 kick 【样例4】 kun nichia tapu ushi
提示
样例1解释
beet and vet are written on two sheets each, while beat, bed, and bet are written on one vote each. Thus, we should print the strings beet and vet.
题意比较简单，就是输出出现次数较多的字符串，如果有多个，按照字典序来解决
下面附上本蒟蒻的收获
当时清晰地记得这个题在原网站上做过，并且能够顺利通过，当把代码重构再次写出来时，在原网站能够正确通过，但是在UPCoj上却无法通过，后来知道这个题卡了cin cout，根据同学介绍，不得不换为字符串数组来解决，并能顺利卡过
#pragma GCC optimize (2) #pragma G++ optimize (2) #include <bits/stdc++.h> #include <algorithm> #include <map> #include <queue> #include <set> #include <stack> #include <string> #include <vector> using namespace std; #define wuyt main typedef long long ll; #define HEAP(...) priority_queue<__VA_ARGS__ > #define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > > template<class T> inline T min(T &x,const T &y){return x>y?y:x;} template<class T> inline T max(T &x,const T &y){return x<y?y:x;} //#define getchar()(p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++) //char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf; ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar(); if(c == '-')Nig = -1,c = getchar(); while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar(); return Nig*x;} #define read read() const ll inf = 1e15; const int maxn = 1e6 + 7; const int mod = 1e9 + 7; int num[maxn]; map<string,ll>mp; map<string,ll>::iterator it; string ss[maxn]; int main() { /** int n=read; string ss; map<string,int>mp; for(int i=0;i<n;i++) { ///scanf("%s",ss); cin>>ss; mp[ss]++; } int cnt=0; for(auto& v:mp) cnt=max(cnt,v.second); for(auto& v:mp){ if(v.second==cnt) ///cout<<v.first<<endl; printf("%s\n",v.first); } ll n; ios_base::sync_with_stdio(false); cin>>n; for(ll i=0;i<n;i++) { ///scanf("%s",ss); ///getchar(); string ss; cin>>ss; mp[ss]++; } int maxx=-1; for(const auto& x : mp){ int temp=x.second; if(temp>maxx) maxx=temp; } for(auto it=mp.begin();it!=mp.end();it++){ if(it->second==maxx) ///cout<<it->first<<endl; printf("%s\n",it->first); } int n; cin>>n; int maxx=0; for(int i=1;i<=n;i++) { string s; ///scanf("%s",&s); cin>>s; mp[s]++; maxx=max(maxx,mp[s]); } for(auto x:mp) if(x.second==maxx) cout<<x.first<<endl;**/ ll n=read; ll maxx=0; for(ll i=1;i<=n;i++){ cin>>ss[i]; mp[ss[i]]++; } for(it=mp.begin();it!=mp.end();it++) maxx=max(maxx,it->second); sort(ss+1,ss+1+n); for(ll i=1;i<=n;i++){ if(mp[ss[i]]!=maxx) continue; ll len=ss[i].length(); for(ll j=0;j<len;j++) putchar(ss[i][j]); printf("\n"); mp[ss[i]]=0; } return 0; } /**ll kruskal(){ sort(num,num+m,cmp); for(ll i=1;i<=n;i++) num2[i]=i; ll res=0,cnt=0; for(ll i=0;i<m;i++){ ll aa=num[i].a,b=num[i].b,w=num[i].w; aa=searchnum(aa),b=searchnum(b); if(aa!=b){ num2[aa]=b; res+=w; cnt++; } } if(cnt<n-1) return inf; else return res; }**/ /************************************************************** Language: C++ Result: 正确 Time:659 ms Memory:52768 kb ****************************************************************/
在前面已经附上了若干版本
但是卡不过去
当时还发生了玄学问题：
加上这句之后，不能AC
ios_base::sync_with_stdio(false);
知道的同志们可以在下方评论告诉本蒟蒻
