PTA乙级题 1080. MOOC期终成绩 (25)

【题目链接】

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最后一个点运行超时，询问大佬后得知，匹配名字处（即41行和55行处）过于复杂，第一个代码容易理解，但运行超时，第二个代码AC了。

#include<stdio.h> #include<string.h> typedef struct node { char id[25]; int a; int b; int c; int sum; }Node; int main() { int x=0,score,i,j,q,w,e; char ID[25]; scanf("%d %d %d",&q,&w,&e); getchar(); Node s[q],temp; //建立数组 for (i=0;i<q;i++) { memset(ID,0,sizeof(ID)); scanf("%s %d",ID,&score); if (score>=200) { strcpy(s[x].id,ID); s[x].a=score; x++; } } //设置默认值 for (i=0;i<x;i++) { s[i].b=-1; s[i].c=-1; } //读取期中成绩 for (i=0;i<w;i++) { memset(ID,0,sizeof(ID)); scanf("%s %d",ID,&score); for (j=0;j<x;j++) { if (strcmp(ID,s[j].id)==0) { s[j].b=score; break; } } } //读取期末成绩 for (i=0;i<e;i++) { memset(ID,0,sizeof(ID)); scanf("%s %d",ID,&score); for (j=0;j<x;j++) { if (strcmp(ID,s[j].id)==0) { s[j].c=score; break; } } } //计算总成绩 for (i=0;i<x;i++) { if (s[i].b>s[i].c) s[i].sum=s[i].b*0.4+s[i].c*0.6+0.5; //此处应该+0.5，四舍五入。 else s[i].sum=s[i].c; } //排序 for (i=0;i<x-1;i++) { for (j=0;j<x-1-i;j++) { if (s[j].sum<s[j+1].sum||(s[j].sum==s[j+1].sum&&strcmp(s[j].id,s[j+1].id)>0)) { temp=s[j]; s[j]=s[j+1]; s[j+1]=temp; } } } for (i=0;i<x;i++) { if (s[i].sum>=60) printf("%s %d %d %d %d\n",s[i].id,s[i].a,s[i].b,s[i].c,s[i].sum); } }

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下面是芥末三文鱼提供的满分C++代码，有兴趣的同学可以研究一下。

#include <iostream> #include <string> #include <map> #include <vector> #include <algorithm> #include <utility> #include <array> using namespace std; int main() { ios::sync_with_stdio(false); size_t counts[3]; map<string, array<int, 4>> m; cin >> counts[0] >> counts[1] >> counts[2]; for (size_t i = 0; i < 3; i++) for (size_t j = 0; j < counts[i]; j++) { string name; array<int, 4> value = { -1, -1, -1 }; cin >> name >> value[i]; if (i == 0) m.emplace(make_pair(name, value)); else { auto result = m.find(name); if (result != m.end()) result->second[i] = value[i]; else m.emplace(make_pair(name, value)); } } vector<pair<string, array<int, 4>>> v(m.begin(), m.end()); for (auto i = v.begin(); i != v.end(); i++) { if (i->second[1] > i->second[2]) { double middle = i->second[1] == -1 ? 0 : i->second[1]; double final = i->second[2] == -1 ? 0 : i->second[2]; double sum = (0.4 * middle) + (0.6 * final); i->second[3] = (int)sum + (sum - (int)sum > 0.5 ? 1 : 0); } else i->second[3] = i->second[2]; } sort(v.begin(), v.end(), [](pair<string, array<int, 4>> a, pair<string, array<int, 4>> b) { if (a.second[3] != b.second[3]) return isgreater(a.second[3], b.second[3]); else return a.first < b.first; }); for (auto i : v) { if (i.second[3] >= 60 && i.second[0] >= 200) { cout << i.first; for (auto j : i.second) cout << " " << j; cout << endl; } } return 0; }