A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 364041 Accepted Submission(s): 70825
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.
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.
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.
Sample Input 2 1 2 112233445566778899 998877665544332211
Sample Output Case 1: 1 + 2 = 3
Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110
思路
1、利用字符串方式存储数据
2、根据字符1与int1的ascll相差48转换
3、按照逐位相加的方法两个数组进行相加（重要）
4、输出
5、注意格式（重要）
C解法
#include <stdio.h>
#include <string.h>
int main()
{
int i,h;
scanf(
"%d",&h);
for(i=
1;i<=h;i++)
{
char s1[
1001],s2[
1001];
int a[
1001]={
0},b[
1001]={
0},c[
1001]={
0};
int len1,len2,len3;
int max;
int q,k;
int m,d;
int l;
scanf(
"%s %s",s1,s2);
len1=strlen(s1);
len2=strlen(s2);
if(len1>len2)
{
max=len1;
}
else
{
max=len2;
}
for(q=
0,k=len1-
1;q<len1;q++,k--)
{
a[q]=s1[k]-
48;
}
for(q=
0,k=len2-
1;q<len2;q++,k--)
{
b[q]=s2[k]-
48;
}
for(d=
0,m=
0;m<max;m++)
{
c[m]=(a[m]+b[m]+d)%
10;
d=(a[m]+b[m]+d)/
10;
}
printf(
"Case %d:\n%s + %s = ",i,s1,s2);
if(d!=
0)
{
c[max]=
1;
for(len3=max;len3>=
0;len3--)
{
printf(
"%d",c[len3]);
}
}
else
{
for(len3=max-
1;len3>=
0;len3--)
{
printf(
"%d",c[len3]);
}
}
if(i!=h)
{
printf(
"\n\n");
}
else
{
printf(
"\n");
}
}
return 0;
}
C++解法（OJ语言选G++）
#include <iostream>
#include <string.h>
using namespace std;
int main()
{
int i,h;
cin>>h;
for(i=
1;i<=h;i++)
{
char s1[
1001],s2[
1001];
int a[
1001]={
0},b[
1001]={
0},c[
1001]={
0};
cin>>s1>>s2;
int len1,len2;
len1=
strlen(s1);
len2=
strlen(s2);
int max;
if(len1>len2)
{
max=len1;
}
else
{
max=len2;
}
for(
int q=
0,m=len1-
1;q<len1;q++,m--)
{
a[q]=s1[m]-
48;
}
for(
int q=
0,m=len2-
1;q<len2;q++,m--)
{
b[q]=s2[m]-
48;
}
int d;
for(
int k=
0,d=
0;k<max;k++)
{
c[k]=(a[k]+b[k]+d)%
10;
d=(a[k]+b[k]+d)/
10;
}
cout<<
"Case"<<
" "<<i<<
":"<<
"\n"<<s1<<
" "<<
"+"<<
" "<<s2<<
" "<<
"="<<
" ";
if(d!=
0)
{
c[max]=
1;
for(
int len3=max;len3>=
0;len3--)
{
cout<<c[len3];
}
}
else
{
for(
int len3=max-
1;len3>=
0;len3--)
{
cout<<c[len3];
}
}
if(i!=h)
{
cout<<
"\n\n";
}
else
{
cout<<
"\n";
}
}
return 0;
}
总结
1、一般来说，大数都是用字符串来储存，一个字符表示一个数位。大数就是位数多，数值大的意思。通常来说c语言里的基本数据类型范围是有限的，如long型的最大只能表示几十亿，几十亿也就11位数字长度而已。如果用100长的数组表示，假设数组一个元素存数字的一位，那么总位数可以达到100位，这是基本数据类型无法表示的。
2、基本类型的数据范围： [signed]int ：-32768—-32767 unsigned int:0—65535 long[int]:-2^31—-（2^31-1) unsigned long [int]:0—(2^32-1) float:10^-37—–10^38 double:10^-307—–10^308 long double:10^-4931—-10^4932
