#include<bits/stdc++.h> using namespace std; struct **{ string name; string jnum; string dnum; }; ** k[11]; int main(){ int n; cin>>n; for(int i=0;i<3*n;i++){ int a,b; string c; cin>>a>>b>>c; if(b2){ k[a-1].name=c; }else if(b3){ k[a-1].jnum=c; }else if(b==4){ k[a-1].dnum=c; } } for(int i=0;i<n;i++){ cout<<i+1<<" "<<k[i].name<<" "<<k[i].jnum<<" "<<k[i].dnum<<endl; } return 0; }

嗨害嗨 #include <iostream> #include <cstdio> using namespace std; struct node{ int id; string name, j, d; } a[15]; int main(){ int n; cin >> n; int id, x; string s; for(int i = 1; i <= n * 3; i++){ cin >> id >> x >> s; if(x == 2) a[id].name = s; else if(x == 3) a[id].j = s; else a[id].d = s; }for(int i = 1; i <= n; i++){ cout << i << ' ' << a[i].name << ' ' << a[i].j << ' ' << a[i].d << endl; } }

思路 先看题哈，发现有一个序号，且序号有序，那么..... 用map替换sort将排序改为查找，接着由于除去了一个序号，只剩了三个字符串，用string数组即可 废话少叙，直接上代码：

#include<bits/stdc++.h> using namespace std; struct per{ int id; string name; string jnum; string dnum; }a[100001]; int main(){ int n; cin>>n; int t1,t2; string t3; for(int i=1;i<=n*3;i++){ cin>>t1>>t2>>t3; a[t1].id=t1; if(t22) a[t1].name=t3; if(t23){ a[t1].jnum=t3; } if(t2==4){ a[t1].dnum=t3; } } for(int i=1;i<=n;i++){ cout<<a[i].id<<" "<<a[i].name<<" "<<a[i].jnum<<" "<<a[i].dnum<<endl; } return 0; }

拼音命名流算是被我玩明白了

#include<iostream>#include<cstdio> using namespace std;struct node{int id;string name,j,d;}a[15];int main(){int n;cin>>n;int id,x;string s;for(int i=1;i<=n*3;i++){cin>>id>>x>>s;if(x==2)a[id].name =s;else if(x == 3)a[id].j=s;else a[id].d=s;}for(int i=1;i<=n;i++){cout<<i<<' '<<a[i].name<<' '<<a[i].j<<' '<<a[i].d<<endl;}return 0;}

#include<bits/stdc++.h> #include<fstream> #include<string> using namespace std; struct stu{ string b,c,d; }; int main(){ stu a[10005]; int b; cin >>b; for(int i=1;i<=3*b;i++){ int n;cin>>n; int m;cin>>m; string k; cin>>k; if(m2) a[n].b=k; if(m3) a[n].c=k; if(m==4) a[n].d=k; } for(int i=1;i<=b;i++){ cout <<i<<' '<<a[i].b<<' '<<a[i].c<<' '<<a[i].d<<endl; } return 0; }

#include<bits/stdc++.h> using namespace std; struct xx{ string name; string jnum; string dnum; }; int n; int main(){ cin >> n; xx k[n]; for(int i=0;i<3*n;i++){ int a,b; string c; cin >> a >> b >> c; if(b2){ k[a-1].name=c; } else if(b3){ k[a-1].jnum=c; } else if(b==4){ k[a-1].dnum=c; } } for(int i=0;i<n;i++){ cout << i+1<< " " << k[i].name << " " << k[i].jnum << " " << k[i].dnum << endl; } } 记得点赞

#include<bits/stdc++.h> using namespace std; struct em{ int id; string b[11]; }a[101]; int main(){ int n; cin>>n; for(int i=1;i<=n*n;i++){ int m,j; cin>>m>>j; cin>>a[m].b[j]; }