#include

#include

int np;

int temp[20];

struct pe

{

char name[20];

int money;

int back;

int fri;

int temp;

char frd[20][20];

}pe[20];

struct pe *search(char name[20])

{

struct pe *t=&pe[0];

while(1)

{

if(!(strcmp(name,t->name))){return t;break;}

t++;

}

}

int give()

{

struct pe *p=&pe[0];

struct pe *tail=&pe[np-1];

struct pe *g;

int i=0;

while(pfri!=0)

{

p->temp=(int)(p->money/p->fri);

p->money=p->money%p->fri;

}

p++;

}

p=&pe[0];

while(pfrd[i]);

g->money+=p->temp;

}

p++;

}

}

int compare()

{

struct pe *p=&pe[0];

struct pe *tail=&pe[np-1];

while(pname,p->money-p->back);

p++;

}

}

int main()

{

char temp[20];

int j,i;

struct pe *p=&pe[0];

scanf("%d",&np);

for(i=0;iname);

for(i=0;imoney,&p->fri);

if(j=p->fri)

{

for(j--;j>=0;j--)

scanf("%s",p->frd[j]);

}

}

for(i=0;i

#include

#include

int main() {

char names[10][15],tname[15];

int np,i,ng,j,present,togive,gp,rp;

int given[10],received[10];

scanf("%d",&np);

for(i=0;i

看懂题目就行了，绝对的基础

初学者以外的可以不做了（提高AC率的随便吧)

C的初学者,贴段C的吧

#include

#include

long np;

char name[10][255];

int find(char st[255])

{

long i;

for(i=1;i

type

gift=

　　record

　　 ans,money,ngi:integer;

　　 ng:array [1..10] of string;

　　 n:string;

　　end;

var

np,i,j,k,m:integer;

s:array [1..10] of gift;

str:string;

begin

for i:=1 to 10 do s[i].ans:=0;

readln(np);

for i:=1 to np do readln(s[i].n);

for i:=1 to np do

　　begin

　　 readln(str);

　　 for j:=1 to np do if s[j].n=str then m:=j;

　　 readln(s[m].money,s[m].ngi);

　　 for j:=1 to s[m].ngi do readln(s[m].ng[j]);

　　end;

for i:=1 to np do if s[i].ngi>0 then

　　begin

　　 s[i].ans:=s[i].ans-s[i].money div s[i].ngi*s[i].ngi;

　　 for j:=1 to s[i].ngi do for k:=1 to np do if s[i].ng[j]=s[k].n then s[k].ans:=s[k].ans+s[i].money div s[i].ngi;

　　end;

for i:=1 to np do writeln(s[i].n,' ',s[i].ans);

end.

TASK: gift1

LANG: PASCAL

Compiling...

Compile: OK

Executing...

Test 1: TEST OK [0.000 secs, 208 KB]

Test 2: TEST OK [0.000 secs, 208 KB]

Test 3: TEST OK [0.000 secs, 212 KB]

Test 4: TEST OK [0.000 secs, 212 KB]

Test 5: TEST OK [0.000 secs, 212 KB]

Test 6: TEST OK [0.000 secs, 208 KB]

Test 7: TEST OK [0.000 secs, 212 KB]

Test 8: TEST OK [0.000 secs, 208 KB]

Test 9: TEST OK [0.000 secs, 208 KB]

All tests OK.

Your program ('gift1') produced all correct answers! This is your

submission #7 for this problem. Congratulations!

Here are the test data inputs:

---|---|- test 1 ---|---|-

10

mitnik

Poulsen

Tanner

Stallman

Ritchie

Baran

Spafford

Farmer

Venema

Linus

mitnik

300 3

Poulsen

Tanner

Baran

Poulsen

1000 1

Tanner

Spafford

2000 9

mitnik

Poulsen

Tanner

Stallman

Ritchie

Baran

Farmer

Venema

Linus

Tanner

1234 1

Poulsen

Stallman

536 3

Farmer

Venema

Linus

Ritchie

2000 1

mitnik

Baran

79 2

Tanner

Farmer

Farmer

0 0

Venema

12 9

mitnik

Poulsen

Tanner

Stallman

Ritchie

Baran

Spafford

Farmer

Linus

Linus

1000 1

mitnik

---|---|- test 2 ---|---|-

5

dave

laura

owen

vick

amr

dave

200 3

laura

owen

vick

owen

500 1

dave

amr

150 2

vick

owen

laura

0 2

amr

vick

vick

0 0

---|---|- test 3 ---|---|-

2

john

lennon

lennon

0 0

john

0 0

---|---|- test 4 ---|---|-

10

Alex

Bob

Catherine

Dave

Ebert

Francis

Godot

Harris

Iliya

Jimbo

Alex

2000 9

Bob

Catherine

Dave

Ebert

Francis

Godot

Harris

Iliya

Jimbo

Bob

2000 9

Alex

Catherine

Dave

Ebert

Francis

Godot

Harris

Iliya

Jimbo

Catherine

2000 9

Alex

Bob

Dave

Ebert

Francis

Godot

Harris

Iliya

Jimbo

Dave

2000 9

Alex

Bob

Catherine

Ebert

Francis

Godot

Harris

Iliya

Jimbo

Ebert

2000 9

Alex

Bob

Catherine

Dave

Francis

Godot

Harris

Iliya

Jimbo

Francis

2000 9

Alex

Bob

Catherine

Dave

Ebert

Godot

Harris

Iliya

Jimbo

Godot

2000 9

Alex

Bob

Catherine

Dave

Ebert

Francis

Harris

Iliya

Jimbo

Harris

2000 9

Alex

Bob

Catherine

Dave

Ebert

Francis

Godot

Iliya

Jimbo

Iliya

2000 9

Alex

Bob

Catherine

Dave

Ebert

Francis

Godot

Harris

Jimbo

Jimbo

2000 9

Alex

Bob

Catherine

Dave

Ebert

Francis

Godot

Harris

Iliya

---|---|- test 5 ---|---|-

4

these

names

are

dumb

dumb

534 3

these

dumb

are

are

351 1

names

these

509 2

dumb

names

names

278 1

dumb

---|---|- test 6 ---|---|-

2

someguy

someotherguy

someotherguy

1500 1

someguy

someguy

500 1

someotherguy

---|---|- test 7 ---|---|-

8

a

b

c

d

e

f

g

h

c

500 4

a

b

d

h

f

290 3

a

b

c

a

489 7

b

c

d

e

f

h

g

g

0 0

e

1789 2

f

h

d

2000 5

a

b

h

f

e

b

192 5

a

c

h

g

d

h

0 2

a

b

---|---|- test 8 ---|---|-

10

paul

stan

mark

doug

fred

bill

hank

rich

mike

john

paul

0 0

john

300 2

paul

stan

stan

1000 1

paul

mark

2000 3

paul

stan

doug

doug

510 2

paul

stan

fred

1560 2

paul

stan

bill

178 2

paul

stan

hank

97 2

paul

stan

rich

1999 2

paul

stan

mike

1531 2

paul

stan

---|---|- test 9 ---|---|-

10

paul

stan

mark

doug

fred

bill

hank

rich

mike

john

paul

1693 6

stan

mark

doug

fred

bill

hank

john

1843 3

hank

mike

paul

stan

1346 9

paul

mark

fred

bill

doug

hank

rich

mike

john

mark

1657 1

paul

doug

1256 9

paul

stan

bill

mark

fred

rich

hank

mike

john

fred

1250 6

paul

stan

bill

rich

john

mike

bill

1999 2

john

mike

hank

2000 8

stan

mark

doug

fred

rich

bill

mike

john

rich

1999 3

paul

bill

hank

mike

1999 5

hank

bill

mark

rich

john

Keep up the good work!

Thanks for your submission!

Prog9873.pas(18,1) Fatal: illegal character "'?" ($E3)

Fatal: Compilation aborted

为什么会这样,我在自己的机子上做都对

也算练习了字符串吧

type

peo=record

n:string;

m:integer;

end;

var

a:array [1..10] of peo;

n,m,b,q,w,e:integer;

r:string;

#include

using namespace std;

int N;

struct people

{

string name;

int give;

int get;

}p[10];

int num(string name)

{

for(int i=0;i>N;

for(i=0;i>p[i].name;

for(i=0;i>name>>give>>n;

if(n!=0) p[num(name)].give=give-give%n;

for(int j=0;j>name;

p[num(name)].get+=give/n;

}

}

for(i=0;i

#include

#include

#include

typedef struct

{char name[20];

int m,g;

int people[11];

}hp1;

int main(int argc, char *argv[])

{int i,np,j,k,m,n,earn,t,a[100]={0};

hp1 f[11];char p[100],s[100];

scanf("%d",&np);

for(i=1;i

i love you

水题 15分钟

var a:array[1..10] of string; 名称

b:array[1..10] of integer; 盈亏

tp:string; 暂存每次输入的人名（指名单中的）

m,n,i,j,tmp,tp1,tp2,tp3:integer;

变量够了。

function chk(x:string):integer;

var i:integer;

begin

for i:=1 to n do if x=a[i] then exit(i);

end;

判断用。。。

就这么多。。。。。。。

type

gift=

record

ans,money,ngi:integer;

ng:array [1..10] of string;

n:string;

end;

var

np,i,j,k,m:integer;

s:array [1..10] of gift;

str:string;

begin

for i:=1 to 10 do s[i].ans:=0;

readln(np);

for i:=1 to np do readln(s[i].n);

for i:=1 to np do

begin

readln(str);

for j:=1 to np do if s[j].n=str then m:=j;

readln(s[m].money,s[m].ngi);

for j:=1 to s[m].ngi do readln(s[m].ng[j]);

end;

for i:=1 to np do if s[i].ngi>0 then

begin

s[i].ans:=s[i].ans-s[i].money div s[i].ngi*s[i].ngi;

for j:=1 to s[i].ngi do for k:=1 to np do if s[i].ng[j]=s[k].n then s[k].ans:=s[k].ans+s[i].money div s[i].ngi;

end;

for i:=1 to np do writeln(s[i].n,' ',s[i].ans);

end.

var

t:string;

a:array[1..10]of record

name:string;

b:array[1..9]of string;

s,get,money,ngi:longint;

end;

np,i,j,k:longint;

begin

readln(np);

for i:=1 to np do readln(a[i].name);

for i:=1 to np do

begin

readln(t);

for j:=1 to np do if a[j].name=t then break;

readln(a[j].money,a[j].ngi);

if a[j].ngi0 then

begin

a[j].money:=a[j].money-(a[j].money-((a[j].money div a[j].ngi)*a[j].ngi));

for k:=1 to a[j].ngi do

readln(a[j].b[k]);

end;

end;

for i:=1 to np do

for j:=1 to a[i].ngi do

for k:=1 to np do

if (a[k].name=a[i].b[j]) then

a[k].get:=a[k].get+(a[i].money div a[i].ngi);

for i:=1 to np do

begin

a[i].s:=(a[i].get-a[i].money);

write(a[i].name,' ');

writeln(a[i].s);

end;

end.

type

gift=

　　record

　　 ans,money,ngi:integer;

　　 ng:array [1..10] of string;

　　 n:string;

　　end;

var

np,i,j,k,m:integer;

s:array [1..10] of gift;

str:string;

begin

for i:=1 to 10 do s[i].ans:=0;

readln(np);

for i:=1 to np do readln(s[i].n);

for i:=1 to np do

　　begin

　　 readln(str);

　　 for j:=1 to np do if s[j].n=str then m:=j;

　　 readln(s[m].money,s[m].ngi);

　　 for j:=1 to s[m].ngi do readln(s[m].ng[j]);

　　end;

for i:=1 to np do if s[i].ngi>0 then

　　begin

　　 s[i].ans:=s[i].ans-s[i].money div s[i].ngi*s[i].ngi;

　　 for j:=1 to s[i].ngi do for k:=1 to np do if s[i].ng[j]=s[k].n then s[k].ans:=s[k].ans+s[i].money div s[i].ngi;

　　end;

for i:=1 to np do writeln(s[i].n,' ',s[i].ans);

end.

好简单的,别自己想复杂了

函数里忘打局部变量i了，3次WA,愁死我

type

gift=

record

ans,money,ngi:integer;

ng:array [1..10] of string;

n:string;

end;

var

np,i,j,k,m:integer;

s:array [1..10] of gift;

str:string;

begin

for i:=1 to 10 do s[i].ans:=0;

readln(np);

for i:=1 to np do readln(s[i].n);

for i:=1 to np do

begin

readln(str);

for j:=1 to np do if s[j].n=str then m:=j;

readln(s[m].money,s[m].ngi);

for j:=1 to s[m].ngi do readln(s[m].ng[j]);

end;

for i:=1 to np do if s[i].ngi>0 then

begin

s[i].ans:=s[i].ans-s[i].money div s[i].ngi*s[i].ngi;

for j:=1 to s[i].ngi do for k:=1 to np do if s[i].ng[j]=s[k].n then s[k].ans:=s[k].ans+s[i].money div s[i].ngi;

end;

for i:=1 to np do writeln(s[i].n,' ',s[i].ans);

end.

usaco 1.1.2……

删掉文件读入读出直接copy……

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

├ 测试数据 04：答案正确... 0ms

├ 测试数据 05：答案正确... 0ms

├ 测试数据 06：答案正确... 0ms

├ 测试数据 07：答案正确... 0ms

├ 测试数据 08：答案正确... 0ms

├ 测试数据 09：答案正确... 0ms

├ 测试数据 10：答案正确... 0ms

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Accepted 有效得分：100 有效耗时：0ms