type

rec=record

name:string;

qm:longint;

bj:longint;

xg:boolean;

xb:boolean;

lw:boolean;

so:longint;

end;

var

r:array [1..1000] of rec;

max:rec;

n,i,j:integer;

m:longint;

str:string;

begin

readln(n);

for i:=1 to n do

begin

readln(str);

r[i].name:=copy(str,1,pos(' ',str)-1);

delete(str,1,pos(' ',str));

while str[1]' ' do begin r[i].qm:=r[i].qm*10+ord(str[1])-ord('0');delete(str,1,1);end;

delete(str,1,1);

while str[1]' ' do begin r[i].bj:=r[i].bj*10+ord(str[1])-ord('0');delete(str,1,1);end;

delete(str,1,1);

if str[1]='Y' then r[i].xg:=true else r[i].xg:=false;

if str[3]='Y' then r[i].xb:=true else r[i].xb:=false;

delete(str,1,4);

if (ord(str[1])-ord('0')>=1) then r[i].lw:=true else r[i].lw:=false;

r[i].so:=0;

end;

m:=0;

max:=r[1];

for i:=1 to n do

begin

r[i].so:=0;

if (r[i].bj>80) and (r[i].xg) then inc(r[i].so,850);

if (r[i].qm>85) and (r[i].xb) then inc(r[i].so,1000);

if r[i].qm>90 then inc(r[i].so,2000);

if (r[i].qm>85) and (r[i].bj>80) then inc(r[i].so,4000);

if (r[i].qm>80) and (r[i].lw) then inc(r[i].so,8000);

if (r[i].so)>max.so then max:=r[i];

inc(m,r[i].so);

end;

writeln(max.name);

writeln(max.so);

writeln(m);

end.

用记录数组不就OK了....麻烦个啥

program vp1001;

type

student=record

name:string[20];

sorce:integer;

pi:integer;

xg:string;

xs:string;

lw:integer;

jj:integer

end;

var

st:array [1..100] of student;

i,n,max,j,t:integer;

begin

read(n);

for i:=1 to n do

begin

readln(st[i].name);

readln(st[i].sorce,st[i].pi);

readln(st[i].xg);

readln(st[i].xs);

readln(st[i].lw)

end;

{with st[i] do

begin

readln(name);

readln(sorce,pi);

readln(xg);

readln(xs);

readln(lw)

end; }

for i:=1 to n do

st[i].jj:=0;

for i:=1 to n do

begin

if (st[i].sorce>80) and (st[i].lw>=1)

then

st[i].jj:=8000+st[i].jj;

if (st[i].sorce>85) and (st[i].pi>80)

then

st[i].jj:=4000+st[i].jj;

if st[i].sorce>90

then

st[i].jj:=st[i].jj+2000;

if (st[i].sorce>85) and (st[i].xs='y')

then

st[i].jj:=st[i].jj+1000;

if (st[i].pi>80) and (st[i].xs='y')

then

st[i].jj:=st[i].jj+850

end;

for i:=1 to n-1 do

for j:=2 to n do

if st[i].jj>st[j].jj

then

begin

t:=st[j].jj;

st[j].jj:=st[i].jj;

st[i].jj:=t

end;

writeln(st[n].name);

writeln(st[n].jj);

max:=0;

for i:=1 to n do

max:=max+st[i].jj;

writeln(max)

end.

真是搞不懂 如此天书式解题竟然通不过

这里涉及到的就是记录不用也可以，输入数据的字符串处理，求最大值。

Program P1001;

Type rec=record

all,name,w,c:string;

s,e,p:integer;

end;

Var

n,i,j,maxN,money,total:longint;

cop,maxP:string;

a:array[1..100] of rec;

Begin

maxN:=0;

total:=0;

j:=0;

readln(n);

For i:=1 to n do begin

readln(a[i].all);

　　while pos(' ',a[i].all)0 do begin

　　 j:=j+1;

　　 cop:=copy(a[i].all,1,pos(' ',a[i].all)-1);

　　 case j of

　　 1: a[i].name:=cop;

　　 2: val(cop,a[i].s);

　　 3: val(cop,a[i].e);

　　 4: a[i].c:=cop;

　　 5: a[i].w:=cop;

　　 end;

　　 delete(a[i].all,1,pos(' ',a[i].all));

　　if pos(' ',a[i].all)=0 then val(a[i].all,a[i].p);

　　end;

　　j:=0;

If (a[i].s>80) and (a[i].p>=1) then money:=8000;

If (a[i].s>85) and (a[i].e>80) then money:=money+4000;

If (a[i].s>90) then money:=money+2000;

If (a[i].s>85) and ((a[i].w='Y') or (a[i].w='y')) then money:=money+1000;

If (a[i].e>80) and ((a[i].c='Y') or (a[i].c='y')) then money:=money+850;

　　If (MaxN=0) or (money>MaxN) then begin

　　MaxN:=money;

　　MaxP:=a[i].name;

　　end;

Total:= Total + money ;

money:=0;

End;

Writeln(Maxp);

Writeln(Maxn);

Writeln(total);

End.

#include

using namespace std;

struct stu

{

char name[25];

int final;

int group;

char leader;

char west;

int acd;

int scol;

};

stu scolar[100];

int main()

{

int n;

cin>>n;

stu max;

max.scol=-1;

double total=0;

int i;

for(i=0;i>scolar[i].name>>scolar[i].final>>scolar[i].group>>scolar[i].leader>>scolar[i].west>>scolar[i].acd;

scolar[i].scol=0;

if(scolar[i].final>80&&scolar[i].acd>=1)

{

scolar[i].scol+=8000;

}

if(scolar[i].final>85&&scolar[i].group>80)

scolar[i].scol+=4000;

if(scolar[i].final>90)

scolar[i].scol+=2000;

if(scolar[i].final>85&&scolar[i].west=='Y')

scolar[i].scol+=1000;

if(scolar[i].group>80&&scolar[i].leader=='Y')

scolar[i].scol+=850;

if(scolar[i].scol>max.scol)

{

max.scol=scolar[i].scol;

strcpy(max.name,scolar[i].name);

}

total+=scolar[i].scol;

}

printf("%s\n%d\n%.0lf",max.name,max.scol,total);

}

编译通过...

├ 测试数据 01：答案正确... 0ms

├ 测试数据 02：答案正确... 0ms

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├ 测试数据 09：答案正确... 0ms

├ 测试数据 10：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Accepted 有效得分：100 有效耗时：0ms

注意如果最大值不唯一只输出第一个

汗~~

死在这里了

编译通过...

├ 测试数据 01：答案正确... 0ms

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├ 测试数据 10：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Accepted 有效得分：100 有效耗时：0ms

type rec=record

name:string;

m1,m2,sum:longint;

c1,c2:char;

m:byte;

end;

var

a:array[1..100] of rec;

i,j,k,n,ans,max:longint;

c:char;

begin

readln(n);

for i:=1 to n do begin

read(c);

　　　　　　　　while c' ' do begin

　　　　　　　　　　　　a[i].name:=a[i].name+c;

　　　　　　　　　　　　read(c)

　　　　　　　　end;

read(a[i].m1);

read(a[i].m2);

read(c);

read(a[i].c1);

read(c);

read(a[i].c2);

read(a[i].m);

readln;

end;

for i:=1 to n do begin

if (a[i].m1>80) and (a[i].m>0) then begin

a[i].sum:=a[i].sum+8000;

ans:=ans+8000;

end;

if (a[i].m1>85) and (a[i].m2>80) then begin

a[i].sum:=a[i].sum+4000;

ans:=ans+4000;

end;

if a[i].m1>90 then begin

a[i].sum:=a[i].sum+2000;

ans:=ans+2000;

end;

if (a[i].m1>85) and (a[i].c2='Y') then begin

a[i].sum:=a[i].sum+1000;

ans:=ans+1000;

end;

if (a[i].m2>80) and (a[i].c1='Y') then begin

a[i].sum:=a[i].sum+850;

ans:=ans+850;

end;

end;

max:=0;

for i:=1 to n do

if a[i].sum>max then begin

k:=i;

max:=a[i].sum;

end;

writeln(a[k].name);

writeln(max);

writeln(ans);

end.

Program P1001;

Type rec=record

all,name,w,c:string;

s,e,p:integer;

end;

Var

n,i,j,maxN,money,total:longint;

cop,maxP:string;

a:array[1..100] of rec;

Begin

maxN:=0;

total:=0;

j:=0;

readln(n);

For i:=1 to n do begin

readln(a[i].all);

while pos(' ',a[i].all)0 do begin

j:=j+1;

cop:=copy(a[i].all,1,pos(' ',a[i].all)-1);

case j of

1: a[i].name:=cop;

2: val(cop,a[i].s);

3: val(cop,a[i].e);

4: a[i].c:=cop;

5: a[i].w:=cop;

end;

delete(a[i].all,1,pos(' ',a[i].all));

if pos(' ',a[i].all)=0 then val(a[i].all,a[i].p);

end;

j:=0;

If (a[i].s>80) and (a[i].p>=1) then money:=8000;

If (a[i].s>85) and (a[i].e>80) then money:=money+4000;

If (a[i].s>90) then money:=money+2000;

If (a[i].s>85) and ((a[i].w='Y') or (a[i].w='y')) then money:=money+1000;

If (a[i].e>80) and ((a[i].c='Y') or (a[i].c='y')) then money:=money+850;

If (MaxN=0) or (money>MaxN) then begin

MaxN:=money;

MaxP:=a[i].name;

end;

Total:= Total + money ;

money:=0;

End;

Writeln(Maxp);

Writeln(Maxn);

Writeln(total);

End.

好像我方法有点不一样~~~

type rec=record

name:string;

m1,m2,sum:longint;

c1,c2:char;

m:byte;

end;

var

a:array[1..100] of rec;

i,j,k,n,ans,max:longint;

c:char;

begin

readln(n);

for i:=1 to n do begin

read(c);

while c' ' do begin

a[i].name:=a[i].name+c;

read(c)

end;

read(a[i].m1);

read(a[i].m2);

read(c);

read(a[i].c1);

read(c);

read(a[i].c2);

read(a[i].m);

readln;

end;

for i:=1 to n do begin

if (a[i].m1>80) and (a[i].m>0) then begin

a[i].sum:=a[i].sum+8000;

ans:=ans+8000;

end;

if (a[i].m1>85) and (a[i].m2>80) then begin

a[i].sum:=a[i].sum+4000;

ans:=ans+4000;

end;

if a[i].m1>90 then begin

a[i].sum:=a[i].sum+2000;

ans:=ans+2000;

end;

if (a[i].m1>85) and (a[i].c2='Y') then begin

a[i].sum:=a[i].sum+1000;

ans:=ans+1000;

end;

if (a[i].m2>80) and (a[i].c1='Y') then begin

a[i].sum:=a[i].sum+850;

ans:=ans+850;

end;

end;

max:=0;

for i:=1 to n do

if a[i].sum>max then begin

k:=i;

max:=a[i].sum;

end;

writeln(a[k].name);

writeln(max);

writeln(ans);

end.

天啊，这水题我居然用了半个小时，和小学生一个水平，悲哀啊~~~~~~~~

编译通过...

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Accepted 有效得分：100 有效耗时：0ms

小学生就用记录类型+字符串搞了半个小时AC。

var total,i,j,k,l,m,n,b,c:longint;

s,name:array[0..101] of string;

money,rmb:array[0..101] of longint;

flag1,flag2:array[0..101] of boolean;

temp:string;

a:longint;

procedure qsort(l,r:longint);

var i,j,k,x:longint;

begin

i:=l;j:=r;x:=money[(i+j) div 2];

repeat

while money[i]x do dec(j);

if ij;

if i80) and (a>85) then money[i]:=money[i]+4000;

delete(s[i],1,k);

temp:=s;

if (temp='Y') and (b>80) then money[i]:=money[i]+850;

delete(s[i],1,2);

temp:=s;

if (temp='Y') and (a>85) then money[i]:=money[i]+1000;

delete(s[i],1,2);

temp:=s;

val(temp,c,l);

if (c>=1) and (a>80) then money[i]:=money[i]+8000;

end;

for i:=1 to n do begin

rmb[i]:=money[i];

total:=total+money[i];

end;

qsort(1,n);

for i:=1 to n do

if rmb[i]=money[n] then begin

writeln(name[i]);

writeln(rmb[i]);

writeln(total);

break;

end;

close(input);close(output);

end.

快排+字符串处理

#include

int main(){

int i,j,n,qm,py,lw,prize,max=0;

long total=0;

char a[20],name[20],xb,gb;

scanf("%d",&n);

for(i=1;i80)&&(lw>0)) prize+=8000;

if((qm>85)&&(py>80)) prize+=4000;

if(qm>90) prize+=2000;

if((qm>85)&&(xb=='Y')) prize+=1000;

if((py>80)&&(gb=='Y')) prize+=850;

total+=prize;

if(prize>max){

max=prize;

for(j=0;j

program money;

var w,f,d:char; k,len,n,i,j,temp2,sum:integer; temp1:longint;

a:array [1..100,1..5] of longint;

b:array [1..100,1..2] of char;

c:array [1..100] of string;

temp3,e:string;

procedure se;

begin

if ((a>=80) and (a>0)) then a:=a+8000;

if ((a>=85) and (a>=80)) then a:=a+4000;

if (a>=90) then a:=a+2000;

if ((a>=85) and ((b='Y') or (b='y'))) then a:=a+1000;

if ((a>=80) and ((b='Y')or (b='y'))) then a:=a+850;

end;

begin

fillchar(a,sizeof(a),0);

readln(n);

for i:=1 to n do

begin

repeat

read(w);

if w' ' then c[i]:=c[i]+w;

until w=' ';

readln(a,a,f,b,d,b,a);

a:=i;

end;

for i:=1 to n do se;

for i:=1 to n-1 do

for j:=i+1 to n do

if ((aa))) then

begin

temp1:=a;

a:=a[j,4];

a[j,4]:=temp1;

temp2:=a;

a:=a[j,5];

a[j,5]:=temp2;

temp3:=c[i];

c[i]:=c[j];

c[j]:=temp3;

end;

sum:=0;

for i:=1 to n do sum:=sum+a;

writeln(c[1]);

writeln(a[1,4]);

writeln(sum);

end.

为什么错误啊？

1读入数据2看符合条件的人可以得到得奖学金3输出得最大奖学金的人

这种水题需要的果然就是时间，去年在这卡了6次后GIVE UP，然后VIJOS就挂了。。。今年来一看。。。

编译通过...

├ 测试数据 01：答案正确... 0ms

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

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Accepted 有效得分：100 有效耗时：0ms

5) 班级贡献奖，每人850元，班级评议成绩高于80分（>80）的学生干部均可获得；

这里的班级评议成绩看成平均成绩了。。。靠，一道真正做了一年的题就这样解决了。

我是学C语言的，编程时要不要用文件？

#include

using namespace std;

struct abc

{

char name[20];

int a;

int b;

char gb;

char xb;

int c;

long int money;

};

int main()

{

int n;

coutn;

abc *pn=new abc[n];

long int sum=0;

for(int i=0;i>pn[i].name>>pn[i].a>>pn[i].b>>pn[i].gb>>pn[i].xb>>pn[i].c;

sum=0;

if(pn[i].a>80&&pn[i].c>=1)//院士级

sum=sum+8000;

if(pn[i].a>85&&pn[i].b>80)//54级

sum=sum+4000;

if(pn[i].a>90)//成绩优秀

sum=sum+2000;

if(pn[i].a>85&&pn[i].xb=='Y')//西部奖

sum=sum+1000;

if(pn[i].b>80&&pn[i].gb=='Y')//班级

sum=sum+850;

pn[i].money=sum;

}

long int ko=pn[n-1].money;

int k=0;

for(int j=n-1;j>=0;j--)// 比较谁得到最多奖金，顺序倒过来

{

if(pn[j].money>ko)

{

ko=pn[j].money;

k=j;

}

}

long int sumk=0;

for(i=0;i

……这道题交了2遍，原来使用longint，晕，读入的时候我是边读边删的，这道题出了读入，其他都很简单