数字的反转：
就是将数字倒着存下来而已。(*^__^*) 嘻嘻……
大致思路：将数字一位一位取出来，存在一个数组里面，然后再将其变成数字，输出。
详见代码。
while (a) //将每位数字取出来，取完为止
{
num1[i]=a%; //将每一个各位取出存在数组里面，实现了将数字反转
i++; //数组的变化
a/=;
}
趁热打铁 例题:hdu 4554 叛逆的小明
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4554
叛逆的小明
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 818 Accepted Submission(s):
568
Problem Description
叛逆期的小明什么都喜欢反着做，连看数字也是如此（负号除外），比如：
小明会把1234它看成4321；把-1234看成-4321；把230看成032
(032=32)；把-230看成-032(-032=-32)。
小明会把1234它看成4321；把-1234看成-4321；把230看成032
(032=32)；把-230看成-032(-032=-32)。
现在，小明做了一些a+b和a-b的题目(a,
b为整数且不含前导0)，如果给你这些题目正确的答案，你能猜出小明会做得到什么答案吗？
Input
输入第一行为一个正整数T(T<=10000)，表示小明共做了T道题。
接下来T行，每行是两个整数x,y(-1000000<=x,
y<=1000000), x表示a+b的正确答案，y表示a-b的正确答案。
输入保证合法，且不需考虑a或b是小数的情况。
接下来T行，每行是两个整数x,y(-1000000<=x,
y<=1000000), x表示a+b的正确答案，y表示a-b的正确答案。
输入保证合法，且不需考虑a或b是小数的情况。
Output
输出共T行，每行输出两个整数s
t，之间用一个空格分开，其中s表示小明将得到的a+b答案，t表示小明将得到的a-b答案。
t，之间用一个空格分开，其中s表示小明将得到的a+b答案，t表示小明将得到的a-b答案。
Sample Input
3
20 6
7 7
-100 -140
Sample Output
38 24
7 7
-19 -23
题目大意：中文题目，清晰明了~
详见代码。
#include <iostream>
#include <cstdio>
#include <cstring> using namespace std; int num1[],num2[]; int main ()
{
int T;
scanf("%d",&T);
while (T--)
{
int x,y,a,b,i=,j=;
int p=,q=;
scanf("%d%d",&x,&y);
memset(num1,,sizeof(num1));
memset(num2,,sizeof(num2));
a=(x+y)/;
b=x-a;
while (a)
{
num1[i]=a%;
i++;
a/=;
}
while (b)
{
num2[j]=b%;
j++;
b/=;
}
for (int k=; k<i; k++) //换成数字
p=num1[k]+p*;
for (int l=; l<j; l++)
q=num2[l]+q*;
printf ("%d %d\n",p+q,p-q);
}
return ;
}
大数加法：
数字很大，假如直接加。很容易超时！所以换一种方法就是直接取出来一位一位加，不过不能换成数字。
int l=l1>l2?l1:l2; //在两个之间取一个最长的
for (int i=; i<l; i++)
{
num3[i]=num1[i]+num2[i]; //一位一位加
if (num3[i-]>&&i>=) //考虑进位的问题，如果大于9就需要进位
{
num3[i]++;
} }
if (num[l-]>) //最后一位的进位问题
{
num[l]++; //仍需要进位
l++; //长度需要加一
}
例题：hdu 1002 A + B Problem II
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1002
A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 238717 Accepted Submission(s):
46010
Problem Description
I have a very simple problem for you. Given two
integers A and B, your job is to calculate the Sum of A + B.
integers A and B, your job is to calculate the Sum of A + B.
Input
The first line of the input contains an integer
T(1<=T<=20) which means the number of test cases. Then T lines follow,
each line consists of two positive integers, A and B. Notice that the integers
are very large, that means you should not process them by using 32-bit integer.
You may assume the length of each integer will not exceed 1000.
T(1<=T<=20) which means the number of test cases. Then T lines follow,
each line consists of two positive integers, A and B. Notice that the integers
are very large, that means you should not process them by using 32-bit integer.
You may assume the length of each integer will not exceed 1000.
Output
For each test case, you should output two lines. The
first line is "Case #:", # means the number of the test case. The second line is
the an equation "A + B = Sum", Sum means the result of A + B. Note there are
some spaces int the equation. Output a blank line between two test
cases.
first line is "Case #:", # means the number of the test case. The second line is
the an equation "A + B = Sum", Sum means the result of A + B. Note there are
some spaces int the equation. Output a blank line between two test
cases.
Sample Input
2
1 2
112233445566778899 998877665544332211
Sample Output
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
参考以上的详解。将数字一位一位取出来，倒过来相加，在输出~
#include <iostream>
#include <cstdio>
#include <cstring> using namespace std; int main ()
{
int n,flag=;
char a[],b[];
int num1[],num2[],ans[];
scanf("%d",&n);
while (n--)
{
int k=,s=;
scanf("%s%s",a,b);
printf ("Case %d:\n",flag++);
int len1=strlen(a);
int len2=strlen(b);
memset(num1,,sizeof(num1));
memset(num2,,sizeof(num2));
memset(ans,,sizeof(ans));
int l=len1>len2?len1:len2;
for (int i=len1-; i>=; i--)
{
num1[k]=a[i]-'';
k++;
}
for (int j=len2-; j>=; j--)
{
num2[s]=b[j]-'';
s++;
}
for (int i=; i<l; i++)
{
ans[i]=num1[i]+num2[i];
if (i>=)
if (ans[i-]>)
{
ans[i]++;
}
//cout<<num1[i]<<" "<<num2[i]<<" "<<ans[i]<<endl;
}
if (ans[l-]>)
{
ans[l]=;
l++;
}
printf ("%s + %s = ",a,b);
for (int i=l-; i>=; i--)
printf("%d",ans[i]%);
if (n)
printf ("\n");
printf ("\n");
}
return ;
}