hermione_wqx @ 2024-07-26 11:10:34
#include<bits/stdc++.h>
using namespace std;
int main() {
string name, bname;
int c, bc, m, bm, y, by;
int n = 0;
cin >> n;
cin >> name >> c >> m >> y;
for (int i = 1; i < n; i++) {
cin >> bname >> bc >> bm >> by;
if (bc + bm + by > c + m + y) {
c = bc;
m = bm;
y = by;
bname = name;
}
}
cout << name << ' ' << c << ' ' << m << ' ' << y;
return 0;
}
by goIdie @ 2024-07-26 11:14:14
第15行应为
name = bname;
求关
by haimingbei @ 2024-07-26 11:14:58
@goIdie 你这好像啥也没改啊
by goIdie @ 2024-07-26 11:15:57
@haimingbei ,改了啊
by haimingbei @ 2024-07-26 11:16:39
@hermione_wqx 用结构体和sort快排就很简单
by haimingbei @ 2024-07-26 11:17:47
@goIdie 没看到(当我没说)
by hermione_wqx @ 2024-07-26 11:18:26
@goIdie @haimingbei 谢谢两位大佬
by hermione_wqx @ 2024-07-26 11:23:50
@goIdie 已AC
by strin2024 @ 2024-08-01 20:51:06
题解:
#include<>
using namespace ——;
struct stu{
string na;
int c,m,e,s;
}st[];
bool cmp(stu a,stu b){
return a.s>b.s;
}
int main(){
int n,t=1;
cin>>n;
for(int i=1;i<=n;i++){
cin>>st[i].na>>st[i].c>>st[i].m>>st[i].e;
st[i].s=st[i].c + st[i].m + st[i].e;
if(st[i].s>st[t].s) t=i;
}
cout<<st[t].na<<' '<<st[t].c<<' '<<st[t].m<<' '<<st[t].e;
return 0;
}
此题解被做手脚,一
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