mapStudent[456] = "student_second"; 遍历!! 之前从hcy学长那里学到了直接用迭代器来遍历,其实也就等于一个for循环,但是map不能轻易用for循环来遍历 比如(for(int i=0;i<map.size();i++)),这里就相当于我们自动默认了map的key为int,当然不对! 于是我们使用迭代器来遍历map map<x,y> some::iterator ...
**思路:**用一个map记录每个字符串的次数,遍历一遍即可~ 代码: #include<bits/stdc++.h>usingnamespacestd;constintmaxn=1e6+7; map<string,int>mp; string s[maxn]; std::map<string,int>::iterator it;boolcmp(string a,string b){returna<b; }intmain(){intn;cin>>n;for(inti=1;i<=n;i...
AtCoder Beginner Contest 216【C:简单思维】【D:双端队列+BFS 模拟】【E:优先队列+map 数学】 1 #include<bits/stdc++.h> 2 3 using namespace std; 4 #define int long long 5 #define pb push_back 6 7 int n; 8 vector<int> v; 9 signed main(){ 10 cin>>n; 11 while(n){ 12 if(n...
#include<bits/stdc++.h>using namespace std;#define endl'\n';voidbest_coder(){int n;cin>>n;unordered_map<string,string>g(n);unordered_map<string,pair<int,int>>vp(n);for(int i=0;i<n;++i){string a,b;cin>>a>>b;g[b]=a;++vp[a].first;++vp[b].second;}queue<string>q;for...
C - Swappable 在数组中找到满足条件的数对\((i,j)\) \(1 \le i < j \le N (N\in[2,3e5])\) \(A_i \not= A_j\) 一道经典利用map减少搜索规模的题, 先假设每个数互不相同:ans = n * (n - 1) / 2 map存每个数出现的次数,然后减去相同的情况ans -= x * (x - 1) / 2 ...
#include<map> #include<iostream> #include<vector> #include<algorithm> using namespace std; int main(){ int n,w; cin >> n >> w; vector<pair<int,int>>event; for(int i=0;i<n;i++){ int s,t,p; cin >> s >> t >> p; event.push_back(make_pair(s,p)); event.push_back...
//cpeditor.org)#include<iostream>#include<cstdio>#include<string>#include<ctime>#include<cmath>#include<cstring>#include<algorithm>#include<stack>#include<climits>#include<queue>#include<map>#include<set>#include<sstream>#include<cassert>#include<bitset>#include<list>#include<unordered_map>//#...
题解: 线段树 #include"stdafx.h"#include<iostream>#include<cstdio>#include<algorithm>#include<map>usingnamespacestd;#defineMAXN(200010)typedefintll;intn,l[MAXN],r[MAXN],r1[MAXN],cnt;map<int,int>idx;structnode{ll val;node(){val=-(1ll<<30);}}t1[MAXN<<2],t2[MAXN<<2];#define...
所以我们用 map 储存所有的f[i] 的个数。 然后每个f[i] 我们从cnt个选2个组成一对 l ,r。遍历map就可以了。 代码: #include<bits/stdc++.h>#define int long longusingnamespacestd;voidxw(){strings1;cin>>s1;intn=s1.size();vector<int>cnt(10);vector<int>f(n+10);f[0]=0;for(inti=0...
void sol(){ map<string,int> mp; string s="ABCDEABCDE"; rep(i,0,s.size()-2) { mp[s.substr(i,2)]=1; } reverse(all(s)); rep(i,0,s.size()-2) { mp[s.substr(i,2)]=1; } string a,b;cin>>a>>b; cout<<(mp.count(a)==mp.count(b)?"Yes":"No"); } ...