Vijos 题解：http://hi.baidu.com/umule/item/2c997f8ed9600fdae596e017
有疑问请留言 共同进步

###代码
var qm,bg,lw,mn,bj:array[1..100] of integer;
i,j,n,y,z:integer;
zongfen:longint;
name,b:string;
xb,gb:array[1..100] of char;
begin
readln(n);
for i:=1 to n do
read(name[i]);
read(qm[i]);
read(bj[i]);
read(gb[i]);
read(xb[i]);
read(lw[i]);
for i:=1 to n do
begin
if (qm[i]>80) and (lw[i]>=1) then mn[i]:=mn[i]+8000;
if (qm[i]>85) and (bj[i]>80) then mn[i]:=mn[i]+4000;
if (qm[i]>90) then mn[i]:=mn[i]+2000;
if (xb[i]='Y') and (qm[i]>85) then mn[i]:=mn[i]+1000;
if (bj[i]>80) then mn[i]:=mn[i]+850;
end;
j:=0;
for i:=1 to n do zongfen:=zongfen+mn[i];
for i:=1 to n do
begin
if mn[i]>j then
begin
j:=mn[i];
b:=name[i];
y:=mn[i];
z:=zongfen;
end;
end;
end.
为什么运行时错误。。。求助

program test;
var
a,b,c,d,n:string;
i,j,k,x,y,z,r,v,s:integer;
t:longint;
begin
readln(k);
for i:=1 to k do
begin
readln(a);
j:=pos(' ',a);
b:=copy(a,1,j-1);
delete(a,1,j);
j:=pos(' ',a);
val(copy(a,1,j-1),x,r);
delete(a,1,j);
j:=pos(' ',a);
val(copy(a,1,j-1),y,r);
delete(a,1,j);
j:=pos(' ',a);
c:=copy(a,1,j-1);
delete(a,1,j);
j:=pos(' ',a);
d:=copy(a,1,j-1);
delete(a,1,j-1);
val(copy(a,1,length(a)),z,r);
if (x>80) and (z>=1) then v:=v+8000;
if (x>85) and (y>80) then v:=v+4000;
if x>90 then v:=v+2000;
if (x>85) and (d='Y') then v:=v+1000;
if (x>80) and (c='Y') then v:=v+850;
if s<v then
begin
s:=v;
n:=b;
end;
t:=t+v;
v:=0;
end;
writeln(n);
writeln(s);
writeln(t);
end.

以此庆祝我的第一个c++程序
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{ char kun,gan;
char name[101][100];
int max,mark,classmark;
int n,wen,jiang[101];
cin>>n;
max=0;
int i;
for (i=1;i<=n;i++)
{
scanf("%s %d %d %c %c %d",name[i],&mark,&classmark,&gan,&kun,&wen);
jiang[i]=0;
if ((mark>80)&&(wen>=1))
{jiang[i]+=8000;}
if ((mark>85)&&(classmark>80))
{jiang[i]+=4000;}
if (mark>90)
{
jiang[i]+=2000;
}
if ((mark>85)&&((int) kun == (int)'Y'))
{
jiang[i]+=1000;
}
if ((classmark>80)&&((int) gan== (int)'Y'))
{
jiang[i]+=850;
}
if (jiang[i]>max) {max=jiang[i];}
}
int sum=0;
for (i=1;i<=n;i++)
{
sum+=jiang[i];
}
for (i=1;i<=n;i++)
{
if (max==jiang[i]){printf("%s\n",name[i]);printf("%d\n",jiang[i]);break;}
}
printf("%d\n",sum);
return 0;
}

#include <cstdio>
#include <cstring>
#include <algorithm>
#define REP( i, n ) for ( int i = 1; i <= n; i ++ )
#define FOR( i, a, b ) for ( int i = a; i <= b; i ++ )
#define DWN( i, a, b ) for ( int i = b; i >= a; i -- )
#define RST( a, x ) memset ( a, x, sizeof ( a ) );
#define QN q[ i ].name
#define QS1 q[ i ].scr1
#define QS2 q[ i ].scr2
#define QT q[ i ].stu
#define QW q[ i ].west
#define QP q[ i ].pap
#define QZ q[ i ].prz
#define QO q[ i ].pos
#define TRUE == 'Y'
#define NSIZE 102
#define ONLINE_JUDGE NULL

using namespace std;
struct arr
{
char name[ 30 ], stu[ 3 ], west[ 3 ];
int scr1, scr2, pap, prz, pos;
};
int n, ans = 0;
arr q[ NSIZE ];
bool cmp ( arr a, arr b )
{
if ( a.prz != b.prz ) return a.prz > b.prz;
return a.pos < b.pos;
}
int main ()
{

#ifndef ONLINE_JUDGE
freopen ( "P1001.in", "r", stdin );
freopen ( "P1001.out", "w", stdout );
#endif

scanf ( "%d", &n );
REP ( i, n )
{
scanf ( "%s%d%d%s%s%d", QN, &QS1, &QS2, QT, QW, &QP );
QZ = 0; QO = i;
if ( QS1 > 80 && QP ) QZ += 8000;
if ( QS1 > 85 && QS2 > 80 ) QZ += 4000;
if ( QS1 > 90 ) QZ += 2000;
if ( QS1 > 85 && QW[ 0 ] TRUE ) QZ += 1000;
if ( QS2 > 80 && QT[ 0 ] TRUE ) QZ += 850;
ans += QZ;
}
sort ( q + 1, q + n + 1, cmp );
printf ( "%s\n%d\n%d\n", q[ 1 ].name, q[ 1 ].prz, ans );
return 0;
}
想不通以前怎么写的……

var i,n,max,m,sum,a1,a2,b:longint;
ch,kong,p1,p2:char;
name,ans:string;
begin
readln(n);
max:=0;
for i:=1 to n do begin
m:=0; name:='';
while true do begin
read(ch);
if ch<>' ' then name:=name+ch else break;
end;
readln(a1,a2,kong,p1,kong,p2,b);
if (a1>80)and(b>0) then m:=m+8000;
if (a1>85)and(a2>80) then m:=m+4000;
if a1>90 then m:=m+2000;
if (a1>85)and(p2='Y') then m:=m+1000;
if (a2>80)and(p1='Y') then m:=m+850;
if m>max then begin
max:=m;
ans:=name;
end;
sum:=sum+m;
end;
writeln(ans);
writeln(max);
writeln(sum);
end.

program p1001;
var name:array[0..100,0..20]of char;
n:integer;
qm,bp,lw:array[0..100]of integer;
gb,xb:array[0..100]of boolean;
jxj:array[0..100]of longint;
num,max:longint;
////////////////////////////////////////////////////////////////////////////////
procedure init;
var ch:char;
i,j,k:integer;
begin
for i:=0 to 100 do
for j:=0 to 20 do
name[i,j]:=' ';
readln(n);
for i:=1 to n do
begin
read(ch);
k:=0;
while ch<>' ' do
begin
inc(k);
name[i,k]:=ch;
read(ch);
end;
read(qm[i],bp[i]);
read(ch);read(ch);
if ch='Y' then gb[i]:=true
else gb[i]:=false;
read(ch);read(ch);
if ch='Y' then xb[i]:=true
else xb[i]:=false;
readln(lw[i]);
end;
end;
////////////////////////////////////////////////////////////////////////////////
procedure main;
var nn,i:integer;
begin
num:=0;max:=0;
fillchar(jxj,sizeof(jxj),0);
for i:=1 to n do
begin
if (qm[i]>80)and(lw[i]>=1) then jxj[i]:=jxj[i]+8000;
if (qm[i]>85)and(bp[i]>80) then jxj[i]:=jxj[i]+4000;
if qm[i]>90 then jxj[i]:=jxj[i]+2000;
if (qm[i]>85)and(xb[i]) then jxj[i]:=jxj[i]+1000;
if (bp[i]>80)and(gb[i]) then jxj[i]:=jxj[i]+850;
if (jxj[i]>max)or(max=0) then
begin
max:=jxj[i];
nn:=i;
end;
num:=num+jxj[i];
end;
for i:=1 to 20 do
begin
if name[nn,i]<>' ' then write(name[nn,i]);
end;
writeln;
writeln(max);
writeln(num);
end;
////////////////////////////////////////////////////////////////////////////////
begin
init;
main;
end.

#include <stdio.h>

#define N 100

int main(){
char name[N][20],of[N],we[N];
int i,n,fe[N],ca[N],ar[N],maxi,maxb,bo[N];
long total;

scanf("%d", &n);
for (i=0; i<n; i++){
scanf("%s %d %d %c %c %d", name[i], &fe[i], &ca[i],\
&of[i], &we[i], &ar[i]);
}
total=0;
maxi=0;
maxb=0;
for (i=0; i<n; i++){
bo[i]=0;
if ((fe[i]>80) && (ar[i]>=1)) bo[i]+=8000;
if ((fe[i]>85) && (ca[i]>80)) bo[i]+=4000;
if (fe[i]>90) bo[i]+=2000;
if ((fe[i]>85) && (we[i]=='Y')) bo[i]+=1000;
if ((ca[i]>80) && (of[i]=='Y')) bo[i]+=850;
total+=bo[i];
if (maxb<bo[i]) {
maxb=bo[i];
maxi=i;
}
}
printf("%s\n%d\n%ld", name[maxi], bo[maxi], total);
return 0;
}

为什么总是40分？求解。

#include <stdio.h>

#define N 100

int main(){
char name[N][20],of[N],we[N];
int i,n,fe[N],ca[N],ar[N],maxi,maxb,bo[N];
long total;

scanf("%d", &n);
for (i=0; i<n; i++){
scanf("%s %d %d %c %c %d", name[i], &fe[i], &ca[i],\
&of[i], &we[i], &ar[i]);
}
total=0;
maxi=0;
maxb=0;
for (i=0; i<n; i++){
bo[i]=0;
if ((fe[i]>80) && (ar[i]>=1)) bo[i]+=8000;
if ((fe[i]>85) && (ca[i]>80)) bo[i]+=4000;
if (fe[i]>90) bo[i]+=2000;
if ((fe[i]>85) && (we[i]=='Y')) bo[i]+=1000;
if ((ca[i]>80) && (of[i]=='Y')) bo[i]+=850;
total+=bo[i];
if (maxb<bo[i]) {
maxb=bo[i];
maxi=i;
}
}
printf("%s\n%d\n%ld", name[maxi], bo[maxi], total);
return 0;
}

为什么总是40分？求解。

为什么总是runtime error？？？

#include<iostream>
using namespace std;
#include<stdio.h>
int main(){
char letter,name[22][101],namemax[21],text[3][101];
int score[4][101],lunwen[101];
int n,i,j,sum=0,max=0;
cin>>n;
for(i=1,j=1;i<=n;i++)
{j=1;
scanf("%c",letter);
for(;letter!=' ';)
{name[i][j++]=letter;
scanf("%c",letter);}

for(j=1;j<=2;j++)
scanf("%d",&score[i][j++]);
scanf("%d",&score[i][j]);
scanf("%d",&lunwen[i]);
if(score[i][1]>80&&lunwen[i]>0)score[i][3]+=8000;
if(score[i][1]>85&&score[i][2]>80)score[i][3]+=4000;
if(score[i][1]>90)score[i][3]+=2000;
if(score[i][1]>85&&text[i][2]=='Y')score[i][3]+=1000;
if(score[i][2]>80&&text[i][1]=='Y')score[i][3]+=850;
sum+=score[i][3];
if(score[i][3]>max)
{max=score[i][3];
for(j=1;j<=20;j++)
namemax[j]=name[i][j];}
}
for(i=1;namemax[i];i++)
printf("%c",namemax[i]);
cout<<' '<<max<<sum;
return 0;
}

第一次写RECORD类型。。。表示压力山大。
主要是 CARDE 和 WEST 的输入比较坑爹 。
以下是AC程序
var i,j,k,l,n,m,ans1,max:longint;
student:array[0..1000] of record
name:string;
score:longint;
classes:longint;
carde:string[2];
west:string[2];
article:longint;
money:longint;
end;
c:char;
ans:string;

begin
readln(n);
for i:=1 to n do
with student[i] do begin
name:='';
c:='a';
while c<>' ' do begin
read(c);
if c<>' ' then name:=name+c;
end;
read(score);
read(classes);
read(carde);
read(west);
readln(article);
if (score>80) and (article>=1) then money:=money+8000;
if (score>85) and (classes>80) then money:=money+4000;
if (score>90) then money:=money+2000;
if (score>85) and (west=' Y') then money:=money+1000;
if (classes>80) and (carde=' Y') then money:=money+850;
if money>max then begin
max:=money;
ans:=name;
end;
ans1:=ans1+money;
end;
writeln(ans);
writeln(max);
writeln(ans1);
end.

来个纯 C 的...

#include <stdio.h>
#include <strings.h>

struct stu_info
{
char name[20];
int avg;
int brownie;
char is_representative;
char is_from_west;
int paper;
};

int calc_scholarship(struct stu_info si);

int main()
{
struct stu_info si;
int num, i = 0, scholarship, max_sch = 0, total = 0;
char max_name[20];

scanf(" %d", &num);

for (; i < num; i++)
{
scanf(" %s %d %d %c %c %d", si.name, &si.avg, &si.brownie, &si.is_representative, &si.is_from_west, &si.paper);

scholarship = calc_scholarship(si);
total += scholarship;

if (scholarship > max_sch)
{
max_sch = scholarship;
strcpy(max_name, si.name);
}
}

printf("%s\n%d\n%d", max_name, max_sch, total);

return 0;
}

int calc_scholarship(struct stu_info si)
{
int scholarship = 0;

if (si.avg > 80 && si.paper >= 1)
scholarship += 8000;

if (si.avg > 85 && si.brownie > 80)
scholarship += 4000;

if (si.avg > 90)
scholarship += 2000;

if (si.avg > 85 && si.is_from_west == 'Y')
scholarship += 1000;

if (si.brownie > 80 && si.is_representative == 'Y')
scholarship += 850;

return scholarship;
}

水题不多说。直接读入之后累加奖学金，然后找最大值即可。如果是PASCAL，那么读入可能会比较麻烦，C++可以直接用scanf读入。

#include <cstdio>
#include <cstring>

struct student
{
char name[30],lead[3],west[3];
int s1,s2,art;
void cls()
{
memset(name,0,sizeof(name));
memset(lead,0,sizeof(lead));
memset(west,0,sizeof(west));
s1=s2=art=0;
}
};

student T;
int max,tot,N,i,tmp;
char ans[30];

int main()
{
scanf("%d",&N);
for (i=0;i<N;i++)
{
T.cls();
tmp=0;
scanf("%s%d%d%s%s%d",T.name,&T.s1,&T.s2,T.lead,T.west,&T.art);
if (T.s1>80&&T.art>=1) tmp+=8000;
if (T.s1>85&&T.s2>80) tmp+=4000;
if (T.s1>90) tmp+=2000;
if (T.s1>85&&T.west[0]=='Y') tmp+=1000;
if (T.s2>80&&T.lead[0]=='Y') tmp+=850;
if (tmp>max)
{
max=tmp;
memset(ans,0,sizeof(ans));
strcpy(ans,T.name);
}
tot+=tmp;
}
printf("%s\n%d\n%d\n",ans,max,tot);
return 0;
}

题解

var
ans,max,total,x,y,z,n,i:longint;
a,b,s,p:string;

begin
readln(n);
total:=0;max:=0;
for i:=1 to n do
begin
ans:=0;
readln(s,x,y,a,b,z);

if (x>=80) and (z>=1) then inc(ans,8000);
if (x>=85) and (y>=80) then inc(ans,4000);
if (x>=90) then inc(ans,2000);
if (x>=85) and (a='Y') then inc(ans,1000);
if (y>=80) and (b='Y') then inc(ans,850);

inc(total,ans);

if ans>max then begin
p:=s;
max:=ans;
end;
end;
writeln(p);
writeln(max);
writeln(total);
readln;
end.
求大神纠错，谢啦

通俗易懂
var
s:array[1..100]of string;
qimo,banji,lunwen,jiangjin:array[1..100]of longint;
ganbu,xibu:array[1..100]of char;
i,j,n,max:longint;
ch:char;
na:string;
sum:longint;
begin
readln(n);
for i:=1 to n do
begin
read(ch);
while ch<>' ' do
begin
s[i]:=s[i]+ch;
read(ch);
end;
read(qimo[i],banji[i]);
read(ch);
read(ch);
ganbu[i]:=ch;
read(ch);
read(ch);
xibu[i]:=ch;
readln(lunwen[i]);
if (qimo[i]>80)and(lunwen[i]>=1) then inc(jiangjin[i],8000);
if (qimo[i]>85)and(banji[i]>80) then inc(jiangjin[i],4000);
if qimo[i]>90 then inc(jiangjin[i],2000);
if (xibu[i]='Y')and(qimo[i]>85) then inc(jiangjin[i],1000);
if (ganbu[i]='Y')and(banji[i]>80) then inc(jiangjin[i],850);
inc(sum,jiangjin[i]);
end;
max:=jiangjin[1];
for i:=1 to n do if max<jiangjin[i] then
begin
max:=jiangjin[i];
na:=s[i];
end;
writeln(na);
writeln(max);
writeln(sum);
end.

各种小细节
var a:array[1..5] of integer;
n,i,j,k,max,m,mm:longint;
nam,ss,s,nn:string;
begin
readln(n);
for i:= 1 to n do begin
m:=0;
readln(s);
s:=s+' ';
k:=0;
nam:='';
for j:= 1 to length(s) do begin
if copy(s,j,1)=' ' then begin
if k<>0 then begin
if ss='Y' then ss:='1';
if ss='N' then ss:='0';
val(ss,a[k]);
k:=k+1;
ss:='';
end;

if k=0 then begin
nam:=ss;
k:=1;
ss:='';
end;
end;

if copy(s,j,1)<>' ' then ss:=ss+copy(s,j,1);

end;

if (a[1]>80) and (a[5]>=1) then m:=m+8000;
if (a[1]>85) and (a[2]>80) then m:=m+4000;
if a[1]>90 then m:=m+2000;
if (a[1]>85) and (a[4]=1) then m:=m+1000;
if (a[2]>80) and (a[3]=1) then m:=m+850;
if m>max then
begin

max:=m;
nn:=nam;
end;
mm:=mm+m;
end;
writeln(nn);
writeln(max);
write(mm);
end.

#include<iostream>
#include<string>
using namespace std;
int main()
{
int maxjj=-1,sumjj=0,jj,n;
string maxjjn,jjn;
int qmpjcj,bjpycj,fblws;
char xsgb,xbsfxs;
cin>>n;
for(int i=0;i<n;i++)
{
cin>>jjn>>qmpjcj>>bjpycj>>xsgb>>xbsfxs>>fblws;
jj=0;
if(qmpjcj>80&&fblws>0)
jj+=8000;
if(qmpjcj>85&&bjpycj>80)
jj+=4000;
if(qmpjcj>90)
jj+=2000;
if(qmpjcj>85&&xbsfxs=='Y')
jj+=1000;
if(bjpycj>80&&xsgb=='Y')
jj+=850;
sumjj+=jj;
if(jj>maxjj)
maxjj=jj,maxjjn=jjn;
}
cout<<maxjjn<<endl<<maxjj<<endl<<sumjj<<endl;
return 0;
}

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;
int main()
{
int n;
int a,b,c,res=0,summ=0;
char mname[11];
char d,e;
scanf("%d",&n);
for(int i=0;i<n;i++)
{
char name[11]="";
int sum=0;
scanf("%s %d %d %c %c %d",name,&a,&b,&d,&e,&c);
if(a>80 && c>=1)
sum+=8000;
if(a>85 && b>80)
sum+=4000;
if(a>90)
sum+=2000;
if(a>85 && e=='Y')
sum+=1000;
if(b>80 && d=='Y')
sum+=850;
res+=sum;
if(sum>summ)
{
for(int j=0;j<=strlen(mname);j++)
mname[j]=0;
for(int j=0;j<=strlen(name);j++)
mname[j]=name[j];
summ=sum;
}
}
puts(mname);
printf("%d\n%d",summ,res);
return 0;
}

题解：http://blog.163.com/syf_light_feather/blog/static/22375506720136288513240/