哎.....又来求助了...C++.....

叮叮发表于 2009-5-21 14:46

/////////////////////////////////////////////////////////////////
// This is a program that reads in big integers, add them up
// and gives out an addition.
// precondition:
//                It will give an erro message when input
//                integer is over 20 digits.
//                It will also give an erro message when the
//                addition is over 20 digits.
/////////////////////////////////////////////////////////////////

#include <iostream>

using namespace std;

//the constent number of digits of an individual input.
const int MAX_NUM = 20;
//the constent number of how many integers will be accepted.
const int MAX = 10;

//A function that add and prints the addition of all the numbers.
//@param the array that stores all the integer
//@param the size of the array.
//@param the numbers of integers were stored.
void count(int,int);

void count(int bigInt[], int x)
{
//initial an array to store the addition.
int result;

//set an integer to store the addition of those digits which are in different
//integer but in the same position.
int sum= 0;
//set an integer to store the extra values.
int up = 0 ;

//a loop that adds all the integers.
for(int i=0; i<=MAX;i++)
{
//a loop that adds up the digits in the same position which
//are in different integers array.
for(int j=0; j < MAX; j++)
{
//test if it`s an available number in that position
if(bigInt<10 && bigInt>=0)
{
//gets the value of those digits in the same position.
sum = sum + bigInt;
}
}
//add the exact value and the carry from the last position together.
int temp = sum%10 +up;
//sight the value into the addition array.
result =temp%10;
//puts the carry value into up and carry it to the next position.
up = temp/10;

}
cout << "The addition of those " << x ;
cout <<" big integers is: " <<endl;
//output the value.
for (int i = 0; i <20 ; i++)
{
if (result != '\n' || result>=0 || result<10)
cout << result;
}
}

int main()
{
//initial an array to store all the numbers.
int bigInt;
//a counter respects to numbers of integers
int x = 0;

cout << "It`s a program that counts the addition of integers."<<endl;
cout << "Please enter your big integer." << endl;

//a loop that gets all the input integers.
while (true)
{
cout << "(it should be positive and less than 20 digits with no blank)"<< endl;

//a counter respects to numbers of digits.
int y = 0;
//set a charactor for cin.
char ch;

//a loop that gets digits for a integer.
while (true)
{
//cin digits as charator.
cin.get(ch);

//test to see if he finish an integer input or the input is wrong.
if(ch== '\n' || ch == '\t')
break;
else if (y >19)
{
cout << "Sorry, you have more than 20 digits, it`s too much."<<endl;
         break;
}
//fits the charactor in integer and stores into the array.
else
bigInt = int(ch)-int('0');

//increase the digits counter.
y++;
}

         //a loop that moves digits in the array to the back.
for(int i=MAX_NUM-1;i >= MAX_NUM-1-y;i--)
{
bigInt=bigInt;
y--;
}

//ask for if any following integers are need to be add.
char ans;
cout << "Would you like to start to enter another big int?(y/n)"<< endl;
cin >> ans;
cout << endl;
//stops the loop when the user say no more integers.
if (ans == 'n' || ans == 'N')
{
cout<<"OK"<<endl;
break;
}
//stops the loop when there`re too many integers.
else if (x >= MAX)
{
cout << "Sorry, you have enter too much integers!"<<endl;
break;
}
//increase the integer counter.
x++;

            cout <<endl;
cout <<x<<endl<<y<<endl;
   }

//call the count function.
count(bigInt,x);

return 0;
}

就是要读取大的整数并且做加法...

我怎么感觉没错呢= =#

PabloPicasso 发表于 2009-5-21 14:53

叮叮发表于 2009-5-21 14:57

cin.get(ch);
....
bigInt = int(ch)-int(\'0\');

这样的表述方式不对吗??

我输入了2..想把它读成\'2\'这个char..然后再用bigInt = int(ch) - int(\'0\')来把它换成整数2存到array里..错了吗?错了吗??

chillicez 发表于 2009-5-21 15:04

没错，写ch-‘0’就可以了

另外，lz你这代码满篇都是int* 到int的赋值错误

叮叮发表于 2009-5-21 15:09

楼上..举个例子啊..

= =####

另外..那个bigInt真的没吃到正确的数字哦...

while (true)
{
//cin digits as charator.
cin.get(ch);

//test to see if he finish an integer input or the input is wrong.
if(ch== \'\\n\' || ch == \'\\t\')
break;
else if (y >19)
{
cout << \"Sorry, you have more than 20 digits, it`s too much.\"<<endl;
break;
}
//fits the charactor in integer and stores into the array.
else if (int(ch)-int(\'0\') >=0 || int(ch)-int(\'0\')<10)
bigInt = int(ch)-int(\'0\');

//increase the digits counter.
y++;
cout<<endl<<\"bigInt is \"<< bigInt << endl << \"y is \" << y <<endl;
}

这个是用来读一个数组的while loop然后那个cout 的结果是些奇怪的东西
1
bigInt is 132008
y is 1
2
bigInt is 127
y is 2
3
bigInt is 0
y is 3
4
bigInt is 2896432
y is 4

叮叮发表于 2009-5-21 15:11

原帖由 PabloPicasso 于 2009-5-21 14:53 发表 http://bbs.saraba1st.com/images/common/back.gif
unsigned int的最大取值范围不超过10位的2^32

你这超过20位才报警

数据类型应该是long long吧

不许用long..这个烂program就是要求做一些array来实现char 和int 的转换做加减法..好像很弱智..但是我还是做砸了...恼火中

chillicez 发表于 2009-5-21 15:18

------ Build started: Project: ttt, Configuration: Debug Win32 ------
Compiling...
ttt.cpp
e:\\mycode\\update\\ttt\\ttt.cpp(45) : error C2446: \'<\' : no conversion from \'int\' to \'int *\'
   Conversion from integral type to pointer type requires reinterpret_cast, C-style cast or function-style cast
e:\\mycode\\update\\ttt\\ttt.cpp(45) : error C2040: \'<\' : \'int \' differs in levels of indirection from \'int\'
e:\\mycode\\update\\ttt\\ttt.cpp(48) : error C2440: \'=\' : cannot convert from \'int *\' to \'int\'
   There is no context in which this conversion is possible
e:\\mycode\\update\\ttt\\ttt.cpp(54) : error C2440: \'=\' : cannot convert from \'int\' to \'int \'
   There are no conversions to array types, although there are conversions to references or pointers to arrays
e:\\mycode\\update\\ttt\\ttt.cpp(64) : error C2446: \'!=\' : no conversion from \'int\' to \'int *\'
   Conversion from integral type to pointer type requires reinterpret_cast, C-style cast or function-style cast
e:\\mycode\\update\\ttt\\ttt.cpp(64) : error C2040: \'!=\' : \'int \' differs in levels of indirection from \'int\'
e:\\mycode\\update\\ttt\\ttt.cpp(64) : error C2446: \'<\' : no conversion from \'int\' to \'int *\'
   Conversion from integral type to pointer type requires reinterpret_cast, C-style cast or function-style cast
e:\\mycode\\update\\ttt\\ttt.cpp(64) : error C2040: \'<\' : \'int \' differs in levels of indirection from \'int\'
e:\\mycode\\update\\ttt\\ttt.cpp(115) : error C2440: \'=\' : cannot convert from \'int\' to \'int \'
   There are no conversions to array types, although there are conversions to references or pointers to arrays
Build log was saved at \"file://e:\\MyCode\\update\\ttt\\Debug\\BuildLog.htm\"
ttt - 9 error(s), 0 warning(s)
========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========

我只是好奇这段代码在什么编译器下可以通过编译

叮叮发表于 2009-5-21 15:23

耶...

在Scite下通过了哦......

我!@#!@#$!@#$!

叮叮发表于 2009-5-21 15:27

呃...

bigInt < 10 && bigInt>=0

bigInt[][]是int 哦..不可以和数字比较吗?

sum = sum + bigInt;这也不行的哦???但是可以compile啊..晕了..

chillicez 发表于 2009-5-21 15:38

什么版本g++呀，即使不加-pedantic你那种情况也绝对会报错的

eph 发表于 2009-5-21 15:46

问题太多了

y++;
和
cout<<endl<<\"bigInt is \"<< bigInt << endl << \"y is \" << y <<endl;
换过来；

int(ch)-int(\'0\') >=0 || int(ch)-int(\'0\')<10
改成 &&

叮叮发表于 2009-5-21 15:48

真的没报哦..

对了对了...现在发现是bigInt[][]读不对数字哦..

我用int tempint = int(ch)-int(\'0\');读那个数字..cout<<tempint 是正确的..但是再用bigInt=tempint输出又不对了..

难道int bigInt[][]不能读数????

应该怎么把数字放到array里啊?我记得以前这么做没错的哦..

另..真的..可以compile..不过pedantic是什么来的哦?

eph 发表于 2009-5-21 15:49

这个问题其实用 256 进制比 10 进制高效得多

eph 发表于 2009-5-21 15:55

void count(int bigInt[], int x)
{
   //initial an array to store the addition.
   int result;

   //set an integer to store the addition of those digits which are in different
   //integer but in the same position.
   int sum = 0;

   //a loop that adds all the integers.
   for(int i = 0; i < MAX_NUM; i++)
   {
                     //a loop that adds up the digits in the same position which
                     //are in different integers array.
                     for(int j = 0; j < x; j++)
                     {
                           //gets the value of those digits in the same position.
                           sum += bigInt;
                     }
                     //add the exact value and the carry from the last position together.
                     //sight the value into the addition array.
                     result = sum % 10;
                     //puts the carry value into up and carry it to the next position.
                     sum /= 10;

   }
   if (sum > 0) cout << \\"...\\" << endl;
   else for (int i = 0; i < MAX_NUM; i++) cout << result;
}

[ 本帖最后由 eph 于 2009-5-21 16:37 编辑 ]

叮叮发表于 2009-5-21 16:04

啊啊啊...楼上的老大...你先解决了我那个array的问题嘛...我..我晕着呢

为什么
int tempint=6;

bigIng = tempint;这样的等式得出的数字不对呢?cout bigInt不是6哦..

eph 发表于 2009-5-21 16:07

回复 15楼的叮叮的帖子

你把 y++ 和 bigInt 的顺序换过来了没

eph 发表于 2009-5-21 16:12

另外你要习惯性地初始化 bigInt 数组为 0

叮叮发表于 2009-5-21 16:14

//add the exact value and the carry from the last position together.
int temp = sum + up;

这个改的不对吧?如果得数大于10的话还是要继续向前进位的...所以sum要变成sum%10的吧?

eph 发表于 2009-5-21 16:16

回复 18楼的叮叮的帖子

你后面既然 temp % 10 了，就不需要 sum % 10 ，否则就丢失十位了。

重新看 14F ，temp 、sum 、up 用一个变量就够了。

[ 本帖最后由 eph 于 2009-5-21 16:18 编辑 ]

叮叮发表于 2009-5-21 16:17

原帖由 eph 于 2009-5-21 16:12 发表 http://bbs.saraba1st.com/images/common/back.gif
另外你要习惯性地初始化 bigInt 数组为 0

初始为0哦......但是我已经用那个IF statement 来判断里面输出的数字啊..为什么还是不行呢...........

崩溃了要..

哦哦哦..我研究下啊...

eph 发表于 2009-5-21 16:19

回复 20楼的叮叮的帖子

你不初始化天知道里边是什么数字

叮叮发表于 2009-5-21 16:34

不是啊...
我现在先这样了...
for (int i = 0; i<=MAX; i++)
{
for (int j = 0; j<=MAX_NUM ; j++)
bigInt = 0;
}

但是 bigInt= int(ch)-int(\'0\');还是读不进数哦..

现在结果是变成了
1
bigInt is 0
x is 0

...救命..
这段代码
bigInt= int(ch)-int(\'0\');

我改成了

int tempint = int(ch)-int(\'0\');
然后我cout <<temp;看了一下..数字是正确的..出来是2..
但是用bigInt=tempint;然后再cout..cout <<bigInt;出来还是0

[ 本帖最后由叮叮于 2009-5-21 16:37 编辑 ]

eph 发表于 2009-5-21 16:40

for (int i = 0; i < MAX; i++)
for (int j = 0; j < MAX_NUM ; j++)
bigInt[i] = 0;
...
bigInt = ch-\'0\';
cout << bigInt;

再不行把整个 main 贴出来

[ 本帖最后由 eph 于 2009-5-21 16:42 编辑 ]

叮叮发表于 2009-5-21 16:44

...我编辑了两次..我打的有的...

但是不知道为什么..好像是被字体吃掉了= =#####

有的...

这个是我的main..

int main()
{
//initial an array to store all the numbers.
int bigInt;

for (int i = 0; i<MAX; i++)
{
for (int j = 0; j<MAX_NUM ; j++)
bigInt[ i ] = 0;
}

//a counter respects to numbers of integers
int x = 0;

cout << \"It`s a program that counts the addition of integers.\"<<endl;
cout << \"Please enter your big integer.\" << endl;

//a loop that gets all the input integers.
while (true)
{
cout << \"(it should be positive and less than 20 digits with no blank)\"<< endl;

//a counter respects to numbers of digits.
int y = 0;
//set a charactor for cin.
char ch;

//a loop that gets digits for a integer.
while (true)
{
//cin digits as charator.
cin.get(ch);

//test to see if he finish an integer input or the input is wrong.
if(ch== \'\\n\' || ch == \'\\t\')
break;
else if (y >19)
{
cout << \"Sorry, you have more than 20 digits, it`s too much.\"<<endl;
         break;
}

//fits the charactor in integer and stores into the array.
bigInt= int(ch)-int(\'0\');

//increase the digits counter.
y++;
cout<<endl<<\"bigInt is \"<< bigInt << endl << \"x is \"<< x << endl <<\" y is \" << y <<endl;
}

         //a loop that moves digits in the array to the back.
for(int i=MAX_NUM-1;i >= MAX_NUM-1-y;i--)
{
bigInt=bigInt;
y--;
}

//ask for if any following integers are need to be add.
char ans;
cout << \"Would you like to start to enter another big int?(y/n)\"<< endl;
cin >> ans;
cout << endl;
//stops the loop when the user say no more integers.
if (ans == \'n\' || ans == \'N\')
{
cout<<\"OK\"<<endl;
break;
}
//stops the loop when there`re too many integers.
else if (x >= MAX)
{
cout << \"Sorry, you have enter too much integers!\"<<endl;
break;
}
//increase the integer counter.
x++;

            cout <<endl;
cout <<x<<endl<<y<<endl;
   }

//call the count function.
count(bigInt,x);

return 0;
}

[ 本帖最后由叮叮于 2009-5-21 16:45 编辑 ]

eph 发表于 2009-5-21 16:44

哦，原来你的 [ i ] 被系统当成斜体了
放在 [ code ] 里边吧

eph 发表于 2009-5-21 16:46

//increase the digits counter.
y++;
cout<<endl<<\\"bigInt is \\"<< bigInt << endl << \\"x is \\"<< x << endl <<\\" y is \\" << y <<endl;你这不是还没换过来么，y 变了，bigInt 当然变了

叮叮发表于 2009-5-21 16:51

啊- -...噢..崩............

会被笑死么..

chillicez 发表于 2009-5-21 16:57

初始化用memset(bigInt, 0, sizeof(int) * MAX * MAX_NUM);if (ans == \'n\' || ans == \'N\')
            {
                     cout<<\\"OK\\"<<endl;
                     break;
            }
            //stops the loop when there`re too many integers.
            else if (x >= MAX)
            {
                     cout << \\"Sorry, you have enter too much integers!\\"<<endl;
                     break;
            }在后面加上 cin.get() 吸收回车，否则你读不进数

x++要写在判断x>=MAX前面，不然可能会溢出             for(int i=MAX_NUM-1;i >= MAX_NUM-1-y;i--)
            {
                     bigInt=bigInt;
                     y--;
            }这种错误我就不说了
我还真是蛋疼

叮叮发表于 2009-5-21 17:01

memset(bigInt, 0, sizeof(int) * MAX * MAX_NUM);

看起来好帅...

哦哦哦..

你不说的错误是什么吖?

深夜2点..脑袋是浆糊做的.....麻烦

叮叮发表于 2009-5-21 17:04

那个cin.get()+在哪里??????回车不是被前面那个CIN>>CH;吃了吗?

ov_efly 发表于 2009-5-21 17:08

y 那个错误明显的
y参加判断循环的条件
你还y-- .......

不会死循环么

chillicez 发表于 2009-5-21 17:20

(y / n)那里输入y后面的回车没被吸收，子循环检测到回车直接退出，所以你读不进数

叮叮发表于 2009-5-21 17:34

原来如此!!!

我X..为什么要手多按回车嘛....真是的...

HyperIris 发表于 2009-5-21 23:50

叮叮发表于 2009-5-22 07:11

好了好了…

其实早上起来…再看一遍…发现自己要多傻有多傻…果然深夜不时干这种事情的时机…

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