var

a,b:integer;

begin

read(a,b);

writeln(a+b);

end.

var

a,b:integer;

begin

readln(a,b);

writeln(a+b);

end.

var

a,b,c:integer;

begin

readln(a,b);

c:=a+b;

writeln(c);

readln;

end.

算法：直接计算

var

opr:array[0..1000000] of char;

data:array[0..1000000] of longint;

str1:string;

l:longint;

procedure init;

begin

readln(str1);

str1:='(' +str1+')';

l:=length(str1);

end;

function count(s1,s2:longint;ch:char):longint;

var

i:longint;

begin

case ch of

'+':count:=s1+s2;

'-':count:=s1-s2;

'*':count:=s1*s2;

'/':count:=s1 div s2;

'^':begin

count:=1;

for i:=1 to s2 do

count:=count*s1;

end;

end;

end;

procedure main;

var

topopr,topdata,i,j,s1,s2,s:longint;

ch:char;

begin

i:=1;

topopr:=0;

topdata:=0;

while i0)

then begin

while (opr[topopr]'(') and (topopr>0) do

begin

if s10 then s1:=data[topdata-1];

if s20 then s2:=data[topdata];

ch:=opr[topopr];

opr[topopr]:=' ';

dec(topopr);

data[topdata]:=0;

dec(topdata);

data[topdata]:=count(s1,s2,ch);

end;

end

else begin

if (opr[topopr]='(') and (topopr1)

then begin

opr[topopr]:=' ';

dec(topopr);

inc(i);

end

else begin

writeln(data[topdata]);

exit;

end;

end;

end;

'+','-': begin

if not (opr[topopr]='(')

then begin

while not (opr[topopr] ='(') do

begin

if s10 then s1:=data[topdata-1];

if s20 then s2:=data[topdata];

ch:=opr[topopr];

opr[topopr]:=' ';

dec(topopr);

data[topdata]:=0;

dec(topdata);

data[topdata]:=count(s1,s2,ch);

end;

inc(topopr);

opr[topopr]:=str1[i];

end

else begin

inc(topopr);

opr[topopr]:=str1[i];

end;

inc(i);

end;

'*','/' :begin

if not (opr[topopr] in ['(','+','-'])

then begin

while not (opr[topopr] in ['(','+','-']) do

begin

if s10 then s1:=data[topdata-1];

if s20 then s2:=data[topdata];

ch:=opr[topopr];

data[topdata]:=0;

opr[topopr]:=' ';

dec(topdata);

dec(topopr);

data[topdata]:=count(s1,s2,ch);

end;

inc(topopr);

opr[topopr]:=str1[i];

end

else begin

inc(topopr);

opr[topopr]:=str1[i];

end;

inc(i);

end;

'^':begin

if opr[topopr]='^'

then begin

while opr[topopr]='^' do

begin

if s10 then s1:=data[topdata-1];

if s20 then s2:=data[topdata];

ch:=opr[topopr];

data[topdata]:=0;

opr[topopr]:=' ';

dec(topdata);

dec(topopr);

data[topdata]:=count(s1,s2,ch);

end;

inc(topopr);

opr[topopr]:=str1[i];

end

else begin

inc(topopr);

opr[topopr]:=str1[i];

end;

inc(i);

end;

'0'..'9':begin

j:=i;

s:=ord(str1[j])-48;

inc(j);

while str1[j] in ['0'..'9'] do

begin

s:=s*10+ord(str1[j])-48;

inc(j);

end;

i:=j;

inc(topdata);

data[topdata]:=s;

end;

end;

end;

end;

begin

init;

main;

end.

高端算法自重

直接读入然后输出a+b即可

vijos这样就AC了，不过额外提一下，某些坑X的OJ[不点名了]需要long long类型，否则会溢出，某些更坑X的OJ有a-b的题目类型，啊为什么这个坑？看看UvaOJ上某道题吧= =至于坑到需要高精度的……似乎还没见到过。。

#include

#include

using namespace std;

int main( )

{   

    int a,b;

    scanf("%d%d",&a,&b);

    int c;

    c=a^b;

    int d;

    int e;

    d=a&b;

#define Plus1 \

    d=d

为什么用cpp写的就一直no cmpiled  fpc就能ac  为什么？

我会说我会用__asm吗（内构汇编）

楼下的楼下的楼下真是够华丽丽的……各种颜色各种错误……；

基础题目，有够基础的……直接加直接输出，pascal什么的都可以直接算表达式输出……

program p;

var

a,b:int64;

begin

read(a,b);

write(a+b)

end.

program p1000;

var a,b:integer;

begin

readln(a,b);

writeln(a+b);

end.

编译通过...

program p1000;

var

a,b:integer;

begin

readln(a,b);

writeln(a+b);

end.

哈哈

program ex;

var x,y:longint;

begin

readln(x,y);

writeln(x+y);

end.

A+B It's So Easy

#include

int main()

{

int a,b;

long sum;

scanf("%d %d",&a,&b);

sum=a+b;

printf("%ld",sum);

return 0;

}

这题还用题解吗？

include

long a;long ps();

void pl()

{exit (0);}

void pp()

{printf("%d\n",ps()+ps());}

long pa()

{scanf("%d",&a);

return (a);}

long ps()

{return (pa());}

int main()

{pp();pl();}

vijos就是比tyvj好啊！

这东西tyvj无法编译