#include <iostream>
#include <cstring>
using namespace std;

struct student
{
string name;
int qmcj;
int pycj;
char gb;
char west;
int lw;
int sum=0;
};

student a[101];
int i, n;
long long sum = 0;
int maxpos, maxl=0;

int main()
{
cin >> n;
for (i = 1; i <= n; i++)
{
cin>>a[i].name>>a[i].qmcj>>a[i].pycj>>a[i].gb>>a[i].west>>a[i].lw;
if (a[i].qmcj > 80 && a[i].lw >= 1)
a[i].sum += 8000;
if (a[i].qmcj > 85 && a[i].pycj > 80)
a[i].sum += 4000;
if (a[i].qmcj > 90)
a[i].sum += 2000;
if (a[i].qmcj > 85 && a[i].west == 'Y')
a[i].sum += 1000;
if (a[i].pycj > 80 && a[i].gb == 'Y')
a[i].sum += 850;
sum += a[i].sum;
if (a[i].sum > maxl)
{
maxpos = i;
maxl=a[i].sum;
}
cin.clear();
}
cout << a[maxpos].name << endl;;
cout << a[maxpos].sum << endl;
cout << sum;
return 0;
}

各位大神求找错
第二个测试点没过
O(∩_∩)O谢谢
type rqq=record
name:string[20];
qm:0..100;
bj:0..100;
gb:char;
xb:char;
lw:0..10;
money:0..15850;
end;
var n,i,b,c:integer;
s:longint;
a:char;
k:array[1..105] of rqq;
begin
readln(n);
for i:=1 to n do
begin
read(a);
while a<>' ' do
begin k[i].name:=k[i].name+a; read(a); end;
read(k[i].qm,k[i].bj);
read(a);
read(k[i].gb);
read(a);
read(k[i].xb,k[i].lw);
end;
for i:=1 to n do
begin
if (k[i].qm>80)and(k[i].lw>0) then k[i].money:=k[i].money+8000;
if (k[i].qm>85)and(k[i].bj>80) then k[i].money:=k[i].money+4000;
if k[i].qm>90 then k[i].money:=k[i].money+2000;
if (k[i].qm>85)and(k[i].xb='Y') then k[i].money:=k[i].money+1000;
if (k[i].bj>80)and(k[i].gb='Y') then k[i].money:=k[i].money+850;
s:=s+k[i].money;
end;
b:=-1;
for i:=1 to n do
if k[i].money>b then
begin b:=k[i].money; c:=i; end;
writeln(k[c].name);
writeln(b);
writeln(s);
end.

一开始以为要开数组，做完了才发现多此一举了……
可以声明一个记录集类型来存所有数据，每行读入每行计算。
要注意读入名字、是否为班干部、是否为西部学生时的数据采集（烦请大神指教）
后附本蒟蒻题解……
###Pascal Code
program p1001;
type
student=record
na:string;
qm,bj,artical,sum:longint;
gb,west:boolean;
end;
var
s:array[1..100] of student;
n,i,j,sum,max,t:longint;
c:char;
begin
for i:=1 to 100 do
begin
s[i].na:='';
s[i].gb:=False;
s[i].west:=False;
s[i].sum:=0;
end;
max:=0; sum:=0;
readln(n);

for i:=1 to n do //read
begin
read(c);
while not (c=' ') do
begin
s[i].na:=s[i].na+c;
read(c);
end;
read(s[i].qm,s[i].bj);
for j:=1 to 4 do
begin
read(c);
if c=' 'then continue;
if (j=2) and (c='Y') then s[i].gb:=True;
if (j=4) and (c='Y') then s[i].west:=True;
end;
readln(s[i].artical); //End read
//Main
t:=0;
if (s[i].qm>80) and (s[i].artical>=1) then inc(t,8000);
if (s[i].qm>85) and (s[i].bj>80) then inc(t,4000);
if s[i].qm>90 then inc(t,2000);
if (s[i].qm>85) and (s[i].west=True) then inc(t,1000);
if (s[i].bj>80) and (s[i].gb=True) then inc(t,850);
if t>s[max].sum then max:=i;
s[i].sum:=t;
sum:=sum+t;
end;
//End Main
writeln(s[max].na); //Write
writeln(s[max].sum);
writeln(sum); //End Write
{for i:=1 to n do //Test
writeln(s[i].na,' ',s[i].qm,' ',s[i].bj,' ',s[i].gb,
' ',s[i].west,' ',s[i].artical);} //End Test
end.

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
char map[100];
int x,y,z;
char w1,w2;
int m;
int ma;
}num[10000];
bool cmp(node a,node b)
{
if(a.m==b.m)
return a.ma<b.ma;
else
return a.m>b.m;
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
int ans=0;
int i,j;
for(i=0;i<n;i++)
{
scanf("%s",num[i].map);
scanf("%d%d",&num[i].x,&num[i].y);
getchar();
scanf("%c",&num[i].w1);
getchar();
scanf("%c",&num[i].w2);
scanf("%d",&num[i].z);
num[i].ma=i;
num[i].m=0;
// printf("%s %d %d %c %c %d\n",num[i].map,num[i].x,num[i].y,num[i].w1,num[i].w2,num[i].z);
}
for(i=0;i<n;i++)
{
if(num[i].x>80&&num[i].z>=1)
{
num[i].m+=8000;
}
if(num[i].x>85&&num[i].y>80)
{
num[i].m+=4000;
}
if(num[i].x>90)
num[i].m+=2000;
if(num[i].x>85&&num[i].w2=='Y')
num[i].m+=1000;
if(num[i].y>80&&num[i].w1=='Y')
num[i].m+=850;
ans+=num[i].m;
}
sort(num,num+n,cmp);
printf("%s\n",num[0].map);
printf("%d",num[0].m);
printf("\n%d\n",ans);
}
return 0;
}

测试数据 #0: Wrong answer, time = 150ms, mem = 1042KiB, score = 0
测试数据 #1: Wrong answer, time = 150ms, mem = 1042 KiB, score = 0
测试数据 #2: Wrong answer, time = 150 ms, mem = 1042 KiB, score = 0
测试数据 #3: Wrong answer, time = 150 ms, mem = 1042 KiB, score = 0
测试数据 #4: Wrong answer, time = 150 ms, mem = 1042 KiB, score = 0
测试数据 #5: Wrong answer, time = 150 ms, mem = 1042 KiB, score = 0
测试数据 #6: Wrong answer, time = 150 ms, mem = 1042 KiB, score = 0
测试数据 #7: Wrong answer, time = 150 ms, mem = 1042 KiB, score = 0
测试数据 #8: Wrong answer, time = 150 ms, mem = 1042 KiB, score = 0
测试数据 #9:Wrong answer, time = 150 ms, mem = 1042 KiB, score = 0
Wrong answer, time = 1500 ms, mem = 1042 KiB, score = 0
var
a,b,c:array[1..100000]of string;
n,i,j,k,p,d,e,f,f2,g:longint;
c1,c2,c3,c4:char;
z1,z2,z3,z4,z5:array[1..1000000]of longint;
bo:boolean;
begin
readln(n);
for i:=1 to n do
readln(a[i]);
p:=0;d:=0; fillchar(z1,sizeof(z1),0); e:=0;
for i:=1 to n do
begin
k:=0;
for j:=1 to length(a[i]) do
while a[i][j]<>' ' do
begin
b[i][j]:=a[i][j];
inc(k);
end;
k:=k+1;
for j:=k to length(a[i]) do
while a[i][j]<>' ' do
begin
p:=p+ord(a[i][j])-48;
inc(k);
end;
k:=k+1;
for j:=k to length(a[i]) do
while a[i][j]<>' ' do
begin
d:=d+ord(a[i][j])-48;
inc(k);
end;
k:=k+1;
for j:=k to length(a[i]) do
while a[i][j]<>' ' do
begin
c1:=a[i][j];
inc(k);
end;
k:=k+1;
for j:=k to length(a[i]) do
while a[i][j]<>' ' do
begin
c2:=a[i][j];
inc(k);
end;
k:=k+1;
for j:=k to length(a[i]) do
g:=ord(a[i][j])-48;
if (p>80) and (g>0) then z1[i]:=z1[i]+8000;
if (p>85) and (d>80) then z1[i]:=z1[i]+4000;
if (p>90) then z1[i]:=z1[i]+2000;
if (p>85) and ((c2='y') or (c2='Y')) then z1[i]:=z1[i]+1000;
if (p>80) and ((c1='y') or (c1='Y')) then z1[i]:=z1[i]+850;
end;
f2:=1;
repeat
bo:=true;
for i:=1 to n-f2 do
if z1[i]<z1[i+1] then
begin
c[i]:=b[i];
b[i]:=b[i+1];
b[i+1]:=c[i];
z2[i]:=z1[i];
z1[i]:=z1[i+1];
z1[i+1]:=z2[i];
end;
until bo;
for i:=1 to n do
e:=e+z1[i];
writeln(b[1]);
writeln(z1[1]);
writeln(e);
end.
无语了

###用面向对象的方法，主要为了练习类与对象的使用
#include <iostream>
#include <string>

using namespace std;
void Proc1();
int main()
{
Proc1();
return 0;
}

class Student
{
public:
Student(string&, short&, short&, char&, char&, short&);
Student();
Student(const Student&);
~Student();
void set_Name(string&);
void set_Final_Score(int&);
void set_Class_Score(int&);
void set_Chairman(char&);
void set_Westman(char&);
string getName();
short Final_Score();
short Class_Score();
char Chairman();
char Westman();
int Scholarship();
private:
string name;
short final_Score,class_Score,paper;
char chairman, westman;
int scholarship;
};

Student::Student()
{
name = "";
final_Score = class_Score = 0;
chairman = westman = 'N';
paper = 0;
scholarship = 0;

}
Student::Student(string&sname, short&FS, short&CS, char&isChairman, char&isWestman, short&npaper)
{
name = sname;
final_Score = FS;
class_Score = CS;
chairman = isChairman;
westman = isWestman;
paper = npaper;
int tmp = 0;
if (final_Score > 80 && paper >= 1) tmp += 8000;
if (final_Score > 85 && class_Score > 80) tmp += 4000;
if (final_Score > 90)tmp += 2000;
if (final_Score > 85 && westman == 'Y')tmp += 1000;
if (class_Score > 80 && chairman == 'Y')tmp += 850;
scholarship = tmp;
}
Student::Student(const Student&Stu)
{
name = Stu.name;
final_Score = Stu.final_Score;
class_Score = Stu.class_Score;
chairman = Stu.chairman;
westman = Stu.westman;
paper = Stu.paper;
scholarship = Stu.scholarship;
}

Student::~Student()
{
}

void Student::set_Name(string& sname)
{
name = sname;
}

void Student::set_Final_Score(int&fs)
{
final_Score = fs;
}
void Student::set_Class_Score(int&cs)
{
class_Score = cs;
}
void Student::set_Chairman(char&isChairman)
{
chairman = isChairman;
}
void Student::set_Westman(char&isWestman)
{
westman = isWestman;
}
string Student::getName()
{
return name;
}
int Student::Scholarship()
{
return scholarship;
}

void Proc1()
{
int n;
cin >> n;
string name_tmp;
short Final_Score, Class_Score,paper;
char isChairman, isWestman;
Student stumax;
int sum = 0;
for (int i = 0; i < n; i++)
{
cin >> name_tmp >> Final_Score >> Class_Score >> isChairman >> isWestman >> paper;
Student stu(name_tmp, Final_Score, Class_Score, isChairman, isWestman, paper);
if (stumax.Scholarship() < stu.Scholarship())
stumax = stu;
sum += stu.Scholarship();
}
cout << stumax.getName() << endl;
cout << stumax.Scholarship() << endl;
cout << sum << endl;

}

果然是因为字符串数量弄错了。。。直接取30个解决。
测试数据 #0: Accepted, time = 2 ms, mem = 496 KiB, score = 10

测试数据 #1: Accepted, time = 2 ms, mem = 492 KiB, score = 10

测试数据 #2: Accepted, time = 0 ms, mem = 492 KiB, score = 10

测试数据 #3: Accepted, time = 0 ms, mem = 496 KiB, score = 10

测试数据 #4: Accepted, time = 0 ms, mem = 492 KiB, score = 10

测试数据 #5: Accepted, time = 0 ms, mem = 496 KiB, score = 10

测试数据 #6: Accepted, time = 0 ms, mem = 492 KiB, score = 10

测试数据 #7: Accepted, time = 0 ms, mem = 492 KiB, score = 10

测试数据 #8: Accepted, time = 2 ms, mem = 496 KiB, score = 10

测试数据 #9: Accepted, time = 0 ms, mem = 496 KiB, score = 10

Accepted, time = 6 ms, mem = 496 KiB, score = 100
代码

#include <stdio.h>
int main()
{
int N, qimo, banji, lunwen, i = 0, I, moneyfinal = 0, moneytotal = 0;
long money;
char ganbu, xibu;
char name[30], namefinal[30];
scanf("%d", &N);
do
{
i = i + 1;
money = 0;
scanf("%s %d %d %c %c %d", &name, &qimo, &banji, &ganbu, &xibu, &lunwen);
if (((qimo > 80) && (lunwen != 0)) == 1) money = money + 8000;
else;
if (((qimo > 85) && (banji > 80)) == 1) money = money + 4000;
else;
if (qimo > 90) money = money + 2000;
else;
if (((qimo > 85) && (xibu == 'Y')) == 1) money = money + 1000;
else;
if (((banji > 80) && (ganbu == 'Y')) == 1) money = money + 850;
else;
moneytotal = moneytotal + money;
if (money > moneyfinal)
{
moneyfinal = money;
for (I = 0; I <= 30; I++) namefinal[I] = name[I];
}
else;
}
while (i < N);
printf("%s\n%d\n%d", namefinal, moneyfinal, moneytotal);
return 0;
}

浪费时间。。。。。好吧很水
#include<stdio.h>
#include<string.h>
char s[25],z[25];
int main()
{
int n,i,a,b,e,qian=0,he=0,zong=0;
char c,d;
scanf("%d",&n);
scanf("%s%d%d %c %c%d",s,&a,&b,&c,&d,&e);
if (a>80&&e>=1) he+=8000;
if (a>85&&b>80) he+=4000;
if (a>90)he+=2000;
if (a>85&&d=='Y') he+=1000;
if (b>80&&c=='Y') he+=850;
strcpy(z,s);
qian=he;
zong=he;
he=0;
for (i=2;i<=n;i++)
{
scanf("%s%d%d %c %c%d",s,&a,&b,&c,&d,&e);
if (a>80&&e>=1) he+=8000;
if (a>85&&b>80) he+=4000;
if (a>90)he+=2000;
if (a>85&&d=='Y') he+=1000;
if (b>80&&c=='Y') he+=850;
if (he>qian)
{qian=he;
strcpy(z,s);}
zong+=he;
he=0;
}
printf("%s\n%d\n%d",z,qian,zong);
return 0;
}

Free Pascal Compiler version 2.6.4 [2014/03/06] for i386
Copyright (c) 1993-2014 by Florian Klaempfl and others
Target OS: Win32 for i386
Compiling foo.pas
Linking foo.exe
85 lines compiled, 0.0 sec , 29472 bytes code, 1628 bytes data
测试数据 #0: Accepted, time = 0 ms, mem = 800 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 804 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 804 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 800 KiB, score = 10
测试数据 #4: Accepted, time = 4 ms, mem = 804 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 800 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 800 KiB, score = 10
测试数据 #7: Accepted, time = 4 ms, mem = 800 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 800 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 804 KiB, score = 10
Accepted, time = 8 ms, mem = 804 KiB, score = 100
代码
##
program p1001;
type t1=record
name:string;
qimo:0..100;
banji:0..100;
ganbu:boolean;
xibu:boolean;
lunwen:boolean;
end;
var a:array[1..100]of t1;
money:array[1..100]of longint;
kong:array[1..5]of longint;
str:string;
n,i,j,k,whole:longint;
function yuanshi(i:longint):boolean;
begin
if (a[i].qimo>80)and a[i].lunwen
then exit(true) else exit(false);
end;
function wusi(i:longint):boolean;
begin
if (a[i].qimo>85)and(a[i].banji>80)
then exit(true) else exit(false);
end;
function youxiu(i:longint):boolean;
begin
if (a[i].qimo>90)
then exit(true) else exit(false);
end;
function xibujiang(i:longint):boolean;
begin
if (a[i].qimo>85)and a[i].xibu
then exit(true) else exit(false);
end;
function banji(i:longint):boolean;
begin
if (a[i].banji>80)and a[i].ganbu
then exit(true) else exit(false);
end;
begin
readln(n);
for i:=1 to n do
begin
readln(str);
k:=0;
for j:=1 to length(str) do
if str[j]=' ' then
begin
inc(k);
kong[k]:=j;
end;
a[i].name:=copy(str,1,kong[1]-1);
val(copy(str,kong[1]+1,kong[2]-kong[1]-1),a[i].qimo,j);
val(copy(str,kong[2]+1,kong[3]-kong[2]-1),a[i].banji,j);
if str[kong[3]+1]='Y'
then a[i].ganbu:=true
else a[i].ganbu:=false;
if str[kong[4]+1]='Y'
then a[i].xibu:=true
else a[i].xibu:=false;
if str[kong[5]+1]='0'
then a[i].lunwen:=false
else a[i].lunwen:=true;
end;
fillchar(money,sizeof(money),0);
whole:=0;
for i:=1 to n do
begin
if yuanshi(i) then inc(money[i],8000);
if wusi(i) then inc(money[i],4000);
if youxiu(i) then inc(money[i],2000);
if xibujiang(i) then inc(money[i],1000);
if banji(i) then inc(money[i],850);
inc(whole,money[i]);
end;
j:=0;k:=0;
for i:=1 to n do
if j<money[i] then
begin
j:=money[i];
k:=i;
end;
writeln(a[k].name);
writeln(money[k]);
writeln(whole);
end.

在vs上亲测可效，不过到了这边一直报错strcyp()没声明

#include<iostream>
#include<string>

using namespace std;

int main()
{
int n, qm, bj, i;
int max = 0,total=0;
char yn1, yn2,Name[20],name[20];
cin >> n;
for (int j = 1; j <= n; j++)
{
int jxj = 0;
cin >> name >> qm >> bj >> yn1 >> yn2 >> i;
if (qm > 80 && i >= 1)
{
jxj = jxj + 8000;
}
if (qm > 85 && bj > 80)
{
jxj = jxj + 4000;
}
if (qm > 90)
{
jxj = jxj + 2000;
}
if (qm > 85 && yn2 == 'Y')
{
jxj = jxj + 1000;
}
if (bj > 80 && yn1 == 'Y')
{
jxj = jxj + 850;
}
total = total + jxj;
if (jxj > max)
{
max = jxj;
strcpy(Name, name);
}

}
cout << Name << endl;
cout << max << endl;
cout<< total << endl;
return 0;
}

简直水
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

string na[105];
char A,B;
bool gb[105],xb[105];
int lw[105],jf[105],pc[105],sum;
int n,i,jxj[105],pm[105],temp,j;

int main()
{
cin>>n;
for (i=1;i<=n;i++)
{
cin>>na[i]>>jf[i]>>pc[i]>>A>>B>>lw[i];
if (A=='N') gb[i]=false;
else gb[i]=true;
if (B=='N') xb[i]=false;
else xb[i]=true;
pm[i]=i;
}
for (i=1;i<=n;i++)
{
if (jf[i]>80 && lw[i]>=1) jxj[i]+=8000; //院士奖学金
if (jf[i]>85 && pc[i]>80) jxj[i]+=4000; //五四奖学金
if (jf[i]>90) jxj[i]+=2000; //成绩优秀奖
if (jf[i]>85 && xb[i]) jxj[i]+=1000; //西部奖学金
if (pc[i]>80 && gb[i]) jxj[i]+=850; //班级贡献奖
sum+=jxj[i];
}
for (i=1;i<n;i++)
for (j=i+1;j<=n;j++)
{
if (jxj[i]<jxj[j]) {
temp=pm[i];pm[i]=pm[j];pm[j]=temp;
temp=jxj[i];jxj[i]=jxj[j];jxj[j]=temp;
}
if (jxj[i]==jxj[j] && pm[i]>pm[j]) {
temp=pm[i];pm[i]=pm[j];pm[j]=temp;
temp=jxj[i];jxj[i]=jxj[j];jxj[j]=temp;
}
}
cout<<na[pm[1]]<<endl<<jxj[1]<<endl<<sum<<endl;
}

#include <cstdio>
#include <cstring>

struct Person
{
char name[21];
int id;
int prize;
} person[100] = {"\0", 0, 0};

int main(void)
{
char name[21] = "\0", job = '\0', west = '\0',
Mnam[21];
int i, n, final, score, article, prize, max = 0, total = 0;

scanf("%d", &n);
for (i = 0; i < n; ++i)
{
scanf("%s %d %d %c %c %d", name, &final, &score, &job, &west, &article);
prize = 0;
if (final > 80)
{
if (article > 0)
{
prize += 8000;
}
if (final > 85)
{
if (score > 80)
{
prize += 4000;
}
if (final > 90)
{
prize += 2000;
}
if (west == 'Y')
{
prize += 1000;
}
}
}
if (score > 80 && job == 'Y')
{
prize += 850;
}
total += prize;
if ((!i) || prize > max)
{
strcpy(Mnam, name);
max = prize;
}
}
printf("%s\n%d\n%d", Mnam, max, total);
return 0;
}

/login be8889f1

#include <iostream>
#include <string>
using namespace std;
int main(){
int a,b=0,c=0;
string d;
cin >> a;
while (a != 0){
string e, f, g;
int h, i, j, k;
cin >> e >> h >> i >> f >> g >> j;
k = (h > 80 && j >= 1 ? 8000 : 0) +
(h > 85 && i > 80 ? 4000 : 0) +
(h > 90 ? 2000 : 0) +
(h > 85 && g == "Y" ? 1000 : 0) +
(i > 80 && f == "Y" ? 850 : 0);
b += k;
if (k > c){
c = k;
d = e;
}
a--;
}
cout << d << endl;
cout << c << endl;
cout << b << endl;;
return 0;
}

#include <iostream>
#include <string>
using namespace std;
int main(){
int 学生的总数,学金的总数=0,获得的奖金最多=0;
string 获得最多奖金的学生的姓名;
cin >> 学生的总数;
while (学生的总数 != 0){
string 姓名, 是否是学生干部, 是否是西部省份学生;
int 期末平均成绩, 班级评议成绩, 发表的论文数, 学金;
cin >> 姓名 >> 期末平均成绩 >> 班级评议成绩 >> 是否是学生干部 >> 是否是西部省份学生 >> 发表的论文数;
学金 = (期末平均成绩 > 80 && 发表的论文数 >= 1 ? 8000 : 0) +
(期末平均成绩 > 85 && 班级评议成绩 > 80 ? 4000 : 0) +
(期末平均成绩 > 90 ? 2000 : 0) +
(期末平均成绩 > 85 && 是否是西部省份学生 == "Y" ? 1000 : 0) +
(班级评议成绩 > 80 && 是否是学生干部 == "Y" ? 850 : 0);
学金的总数 += 学金;
if (学金 > 获得的奖金最多){
获得的奖金最多 = 学金;
获得最多奖金的学生的姓名 = 姓名;
}
学生的总数--;
}
cout << 获得最多奖金的学生的姓名 << endl;
cout << 获得的奖金最多 << endl;
cout << 学金的总数 << endl;
return 0;
}

#include<stdio.h>
int main()
{
int n,i,j,all=0;
scanf("%d",&n);
struct student
{
char a[30],bg,xb;
int qm,by,lw,m;
}stu[n+10];
for(i=1;i<=n;i++)
{
scanf("%s %d %d %c %c %d",stu[i].a,&stu[i].qm,&stu[i].by,&stu[i].bg,&stu[i].xb,&stu[i].lw);
stu[i].m=0;
}

for(i=1;i<=n;i++)
{
if(stu[i].qm>80&&stu[i].lw>=1)
stu[i].m=stu[i].m+8000;
if(stu[i].qm>85&&stu[i].by>80)
stu[i].m=stu[i].m+4000;
if(stu[i].qm>90)
stu[i].m=stu[i].m+2000;
if(stu[i].qm>85&&stu[i].xb=='Y')
stu[i].m=stu[i].m+1000;
if(stu[i].by>80&&stu[i].bg=='Y')
stu[i].m=stu[i].m+850;
}
stu[0]=stu[1];
for(i=1;i<=n;i++)
{
if(stu[0].m<stu[i].m)
stu[0]=stu[i];
all=all+stu[i].m;
}
printf("%s\n%d\n%d",stu[0].a,stu[0].m,all);
return 0;
}

#include"iostream"
#include<cstdio>
#include<cmath>
using namespace std;
typedef long long ll;
const int N=105;
struct node{
char name[21];
int endsc;
int classsc;
char ganbu;
char bibei;
int storys;
int award;
};
int n;
node stu[N];
ll ans;
int maxn=-1;
int k=-1;
void money(node &child,int i)
{
child.award=0;
if(child.endsc>80&&child.storys>=1)
child.award+=8000;
if(child.endsc>85&&child.classsc>80)
child.award+=4000;
if(child.endsc>90)
child.award+=2000;
if(child.endsc>85&&child.bibei=='Y')
child.award+=1000;
if(child.ganbu=='Y'&&child.classsc>80)
child.award+=850;
if(child.award>maxn)
{
maxn=child.award;
k=i;
}
ans+=child.award;
}
int main()
{
cin>>n;
for(int i=0;i<n;i++)
{
cin>>stu[i].name;
cin>>stu[i].endsc;
cin>>stu[i].classsc;
cin>>stu[i].ganbu;
cin>>stu[i].bibei;
cin>>stu[i].storys;

}
for(int i=0;i<n;i++)
money(stu[i],i);
cout<<stu[k].name<<endl;
cout<<maxn<<endl;
cout<<ans;

return 0;
}

测试数据 #0: Accepted, time = 0 ms, mem = 492 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 492 KiB, score = 10
测试数据 #2: Accepted, time = 1 ms, mem = 488 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 492 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 488 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 492 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 488 KiB, score = 10
测试数据 #7: Accepted, time = 15 ms, mem = 492 KiB, score = 10
测试数据 #8: Accepted, time = 2 ms, mem = 492 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 492 KiB, score = 10
#include <cstdio>
#include <cstring>
using namespace std;
int main(){
freopen("input.txt","r",stdin);
char s[100][20],c1,c2;
int n,sum=0,max=0,score1,score2,num,a,b,ans1;
scanf("%d",&n);
for (int i=1;i<=n;i++){
scanf("%s %d %d %c %c %d",&s[i],&score1,&score2,&c1,&c2,&num);
a=c1=='Y'?1:0;
b=c2=='Y'?1:0;
int tmp=0;
if ((score1>80)&&(num>0)) tmp+=8000;
if ((score1>85)&&(score2>80)) tmp+=4000;
if (score1>90) tmp+=2000;
if (b&&(score1>85)) tmp+=1000;
if (a&&(score2>80)) tmp+=850;
sum+=tmp;
if (tmp>max) {max=tmp;ans1=i;}
}
printf("%s\n%d\n%d\n",s[ans1],max,sum);
return 0;

}

刚学c++

水题
program p1001;
var mz,mz1:string;
n,i,qm,bj,lw,max,jxj:integer;
he:longint;
gb,xb:boolean;
c:char;
begin
readln(n);
he:=0;
max:=0;
for i:=1 to n do
begin
jxj:=0;
read(c);
mz:='';
while c<>' ' do
begin
mz:=mz+c;
read(c);
end;
read(qm,bj);
if (qm>85) and (bj>80) then jxj:=jxj+4000;
if qm>90 then jxj:=jxj+2000;
read(c);
read(c);
if c='Y' then gb:=true
else gb:=false;
if (bj>80) and gb then jxj:=jxj+850;
read(c);
read(c);
if c='Y' then xb:=true
else xb:=false;
if (qm>85) and xb then jxj:=jxj+1000;
readln(lw);
if (qm>80) and (lw>0) then jxj:=jxj+8000;
he:=he+jxj;
if jxj>max then
begin
mz1:=mz;
max:=jxj;
end;
end;
writeln(mz1);
writeln(max);
writeln(he);
end.

pascal语言轻松AC
var name:array[1..100] of string;
a1,a2,a5:array[1..100] of longint;
a3,a4:array[1..100] of char;
n,i,max,total,p:longint;
maxname:string;
ch:char;
f:text;
begin
readln(n);
for i:=1 to n do
begin
read(ch);
while ch<>' ' do
begin
name[i]:=name[i]+ch;
read(ch);
end;
readln(a1[i],a2[i],ch,a3[i],ch,a4[i],ch,a5[i]);
end;
for i:=1 to n do
begin
p:=0;
if (a1[i]>80) and (a5[i]>=1) then inc(p,8000);
if (a1[i]>85) and (a2[i]>80) then inc(p,4000);
if (a1[i]>90) then inc(p,2000);
if (a1[i]>85) and (a4[i]='Y') then inc(p,1000);
if (a2[i]>80) and (a3[i]='Y') then inc(p,850);
if p>max then
begin
max:=p;
maxname:=name[i];
end;
inc(total,p);
end;
writeln(maxname);
writeln(max);
writeln(total);
end.