#include

using namespace std;

int main(){

int a,b;

cin>>a>>b;

cout

编译通过...

├ 测试数据 01：Vivid Puppy评测机故障... 32723ms

├ 测试数据 02：Vivid Puppy评测机修理中... 34223ms

├ 测试数据 03：选择其他机器... 3999ms

├ 测试数据 04：Vijos Dragon评测机启动... 1234ms

├ 测试数据 05：紫田漯河双线机房跳电... 1ms

├ 测试数据 06：Vijos Dragon评测机硬件故障... 32323ms

├ 测试数据 07：Vijos Dragon评测机修理中... 32201ms

├ 测试数据 08：Vag 6K评测机启动...4321ms

├ 测试数据 09：Vag 6K评测机安装Free Pascal IDE中... 429421ms

├ 测试数据 10：Vag 6K评测机故障... 142ms

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

unaccepted 有效得分：0 有效耗时：560588ms

// Standard iostream objects -*- C++ -*-

// Copyright (C) 1997, 1998, 1999, 2001, 2002 Free Software Foundation, Inc.

//

// This file is part of the GNU ISO C++ Library. This library is free

// software; you can redistribute it and/or modify it under the

// terms of the GNU General Public License as published by the

// Free Software Foundation; either version 2, or (at your option)

// any later version.

// This library is distributed in the hope that it will be useful,

// but WITHOUT ANY WARRANTY; without even the implied warranty of

// MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the

// GNU General Public License for more details.

// You should have received a copy of the GNU General Public License along

// with this library; see the file COPYING. If not, write to the Free

// Software Foundation, 59 Temple Place - Suite 330, Boston, MA 02111-1307,

// USA.

// As a special exception, you may use this file as part of a free software

// library without restriction. Specifically, if other files instantiate

// templates or use macros or inline functions from this file, or you compile

// this file and link it with other files to produce an executable, this

// file does not by itself cause the resulting executable to be covered by

// the GNU General Public License. This exception does not however

// invalidate any other reasons why the executable file might be covered by

// the GNU General Public License.

//

// ISO C++ 14882: 27.3 Standard iostream objects

//

/** @file iostream

* This is a Standard C++ Library header. You should @c #include this header

* in your programs, rather than any of the "st[dl]_*.h" implementation files.

*/

#ifndef _GLIBCXX_IOSTREAM

#define _GLIBCXX_IOSTREAM 1

#pragma GCC system_header

#include

#include

namespace std

{

/**

* @name Standard Stream Objects

*

* The header declares the eight standard stream

* objects. For other declarations, see

* http://gcc.gnu.org/onlinedocs/libstdc++/27_io/howto.html#10 and the

* @link s27_2_iosfwd I/O forward declarations @endlink

*

* They are required by default to cooperate with the global C library's

* @c FILE streams, and to be available during program startup and

* termination. For more information, see the HOWTO linked to above.

*/

//@{

extern istream cin; ///< Linked to standard input

extern ostream cout; ///< Linked to standard output

extern ostream cerr; ///< Linked to standard error (unbuffered)

extern ostream clog; ///< Linked to standard error (buffered)

#ifdef _GLIBCXX_USE_WCHAR_T

extern wistream wcin; ///< Linked to standard input

extern wostream wcout; ///< Linked to standard output

extern wostream wcerr; ///< Linked to standard error (unbuffered)

extern wostream wclog; ///< Linked to standard error (buffered)

#endif

//@}

// For construction of filebuffers for cout, cin, cerr, clog et. al.

static ios_base::Init __ioinit;

} // namespace std

#endif /* _GLIBCXX_IOSTREAM */

using namespace std;

int main()

{

int a,b;

cin>>a>>b;

cout

#include

using namespace std;

int main(){

int a,b;

cin>>a>>b;

cout

var a,b,c:longint;

begin

read(a,b);

c:=a+b;

write(c);

end.

program sum;

var

x,y,z:integer;

begin

read(x,y);

z:=x+y;

write(z)

end.

.....无语

program exam1;

var a,b,c:integer;

begin

read(a,b);

c:=a+b;

write(c);

end.

var a,b:longint;

begin

readln(a,b);

writeln(a+b);

end.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

这种做法是极其不可取的……

#include

int main(void)

{

int a,b,c;

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

printf("a+b=%d",a+b");

return 0;

}

娃哈哈呀娃哈哈呀，每个人脸上都笑开颜.....

用random然后打表.......

编译通过...

├ 测试数据 01：答案不正确... 999ms

├ 测试数据 02：答案好像正确... 999ms

├ 测试数据 03：答案似乎正确... 999ms

├ 测试数据 04：答案可能正确... 999ms

├ 测试数据 05：答案不能确定... 999ms

├ 测试数据 06：无答案... 999ms

├ 测试数据 07：运行超时... 999ms

├ 测试数据 08：答案失眠... 999ms

├ 测试数据 09：答案休息中... 999ms

├ 测试数据 10：答案不存在... 999ms

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

不能确定 有效得分：0 有效耗时：999牌感冒灵ms

【思路】 by cjf00000

此题考查动态规划思想

首先我们需要把输入字符转换为数字a,b 然后使用动态规划求解A+B的值 最后输出答案

我们可以设计出如下DP方程

令f[j]表示i+j的值 则有

1. f[0][0]=0

2. f[j]=max

3. { f[j]+1,

4. f[i][j-1]+1 }

由于对于每个i,j我们都要计算出f[j],因此时间复杂度与空间复杂度都为O(n^2)

但是 对于题目提供的最大数字n=32767明显超时！

【优化1】by wusu5545

对于DP方程 由于每次计算只需要f[j]或f[j-1]

因此我们可以使用滚动数组优化DP的空间复杂度

使用两个整数x,y分别保存f[j]与f[j-1]

空间复杂度降为O(2) 然而时间复杂度O(n^2)仍然超时!

“时间复杂度，到目前为止还没有更好的优化方法。因此，此题被称为史上最难的dp题!”

【优化2】by 匿名大牛

对于整数的运算 我们可以利用位运算的思想简化复杂度

题目显然要我们求两非负整数之和。

我们知道，在非负整数加法的二进制逻辑运算中，每一位上的结果取决于以下两方面：

1、本位上两个逻辑位的异或值

2、后一位的结果是否溢出

利用这种性质，可以考虑如下做法：

令f[j]表示，考虑两个加数的后i、j位相加的结果，显然有以下状态转移方程

1. f[j]= max

2. { f[i][j-1]+y & (1

var

a,h,i,j,k,l:integer;

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

c,d,e:array[1..100] of integer;

f,g:array[1..100] of char;

m:array[1..101] of longint;

begin

readln(a);

for h:=1 to a do

readln(b[h]);

for h:=1 to a do

begin

i:=1;

b[h]:=b[h]+' ';

while b[h,i]' ' do

i:=i+1;

j:=i-1;

i:=i+1;

while b[h,i]' ' do

begin

c[h]:=c[h]*10+ord(b[h,i])-48;

i:=i+1;

end;

i:=i+1;

while b[h,i]' ' do

begin

d[h]:=d[h]*10+ord(b[h,i])-48;

i:=i+1;

end;

i:=i+1;

f[h]:=b[h,i];

i:=i+2;

g[h]:=b[h,i];

i:=i+2;

while b[h,i]' ' do

begin

e[h]:=e[h]*10+ord(b[h,i])-48;

i:=i+1;

end;

b[h]:=copy(b[h],1,j);

end;

i:=0;

for h:=1 to a do

begin

if (c[h]>80) and (e[h]>0) then

m[h]:=m[h]+8000;

if (c[h]>85) and (d[h]>80) then

m[h]:=m[h]+4000;

if c[h]>90 then

m[h]:=m[h]+2000;

if (c[h]>85) and (g[h]='Y') then

m[h]:=m[h]+1000;

if (d[h]>80) and (f[h]='Y') then

m[h]:=m[h]+850;

m[101]:=m[101]+m[h];

if m[h]>i then

begin

i:=m[h];

j:=h;

end;

end;

writeln(b[j]);

writeln(i);

writeln(m[101]);

end.

这是1001

var

a,b,c:word;

begin

read(a,b);

c:=a+b;

write(c);

end.

终于AC了 交了五次..

我叫张信豪。。谢谢..

program ayqzwu110;

var a,b:longint;

begin

readln(a,b);

writeln(a+b);

end.

program ab;

var a,b:longint;

begin

read(a,b);

write(a+b);

end.

var

a,h,i,j,k,l:integer;

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

c,d,e:array[1..100] of integer;

f,g:array[1..100] of char;

m:array[1..101] of longint;

begin

readln(a);

for h:=1 to a do

readln(b[h]);

for h:=1 to a do

begin

i:=1;

b[h]:=b[h]+' ';

while b[h,i]' ' do

i:=i+1;

j:=i-1;

i:=i+1;

while b[h,i]' ' do

begin

c[h]:=c[h]*10+ord(b[h,i])-48;

i:=i+1;

end;

i:=i+1;

while b[h,i]' ' do

begin

d[h]:=d[h]*10+ord(b[h,i])-48;

i:=i+1;

end;

i:=i+1;

f[h]:=b[h,i];

i:=i+2;

g[h]:=b[h,i];

i:=i+2;

while b[h,i]' ' do

begin

e[h]:=e[h]*10+ord(b[h,i])-48;

i:=i+1;

end;

b[h]:=copy(b[h],1,j);

end;

i:=0;

for h:=1 to a do

begin

if (c[h]>80) and (e[h]>0) then

m[h]:=m[h]+8000;

if (c[h]>85) and (d[h]>80) then

m[h]:=m[h]+4000;

if c[h]>90 then

m[h]:=m[h]+2000;

if (c[h]>85) and (g[h]='Y') then

m[h]:=m[h]+1000;

if (d[h]>80) and (f[h]='Y') then

m[h]:=m[h]+850;

m[101]:=m[101]+m[h];

if m[h]>i then

begin

i:=m[h];

j:=h;

end;

end;

writeln(b[j]);

writeln(i);

writeln(m[101]);

end.

1001的题解!!!!!

program ex_1;

var

a,b:longint;

begin

read(a+b);

writeln(a+b);

readln;

end.

program ex1;

var

a,b:longint;

begin

read(a,b);

write(a+b);

end.

hgdh dfhdfhdfh