[ACM]A + B Problem （大数相加3种方法）
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
解题思路：把两个数的每一位都存放到一个数组中，逆序存放，然后对应位相加的和存放到另一个数组中。
方法1：
#include <iostream>
#include <string.h>
using namespace std;
int main()
{
char x[1001],y[1001];//用来存放输入的两个数，第一个大数存在x里面，第二个大数存在y里面
int xx[1001]={0},yy[1001]={0},sum[1001]={0},i,j=1;//xx数组（整型）用来存放x数组中数字的逆序，因为做加法的时候要从原数的末位开始相加，sum数组用来存放两个大数每位对应相加的值
int lenx,leny,maxlen,t;
cin>>t;
while(t--)
{
cin>>x>>y;
lenx=strlen(x);//x的长度
leny=strlen(y);//y的长度
maxlen=lenx>leny?lenx:leny;//把最大长度赋给maxlen
for(i=0;i<lenx;i++)
xx[i]=x[lenx-i-1]-'0';//根据ascii码表 '0'为48例如 '4'为52 ，相减得到4 ，存放到整型数组xx中，注意x[lenx-i-1]，意思是说逆序存放，为下面逆序相加做准备
for(i=0;i<leny;i++)
yy[i]=y[leny-i-1]-'0';
for(i=0;i<maxlen;i++)
{
sum[i]+=xx[i]+yy[i];//对应位相加
sum[i+1]=sum[i]/10;//因为两个一位数相加不会超过18，如果sum[i]大于10，则sum[i+1]为1，也就是向高位进1
sum[i]=sum[i]%10;//如果对应为相加大于10，取余，相当于 -10；
}
if(sum[i]==1)
maxlen++;//要注意i的值，它代表最高位的后一位，如果最高位大于10大话，要进1，也就是判断s[i]是否为一，进一意味着让位数maxlen+1
cout<<"Case "<<j<<":"<<endl;
j++;
cout<<x<<" + "<<y<<" = ";
for(i=maxlen-1;i>=0;i--)
cout<<sum[i];
cout<<endl;
if(t!=0)
cout<<endl;//没有这一句，AC不了，格式问题
for(i=0;i<maxlen;i++)
{
sum[i]=0;
xx[i]=0;
yy[i]=0;
}//重置，为下一次输入做准备
}
}
方法2：
#include <iostream>
#include <string.h>
using namespace std;
int main()
{
char x[1001],y[10001];
int xx[1001]={0},yy[1001]={0},sum[1001]={0},i,j=1,lenx,leny,maxlen;
int t;
cin>>t;
while(t--)
{
cin>>x>>y;
lenx=strlen(x);
leny=strlen(y);
maxlen=lenx>leny?lenx:leny;
for(i=0;i<lenx;i++)
xx[i]=x[lenx-i-1]-'0';
for(i=0;i<leny;i++)
yy[i]=y[leny-i-1]-'0';
for(i=0;i<maxlen;i++)
{
sum[i]+=xx[i]+yy[i];
sum[i+1]=sum[i]/10;
sum[i]=sum[i]%10;//算法和方法1相同
}
if(sum[i]==1)
maxlen++;
cout<<"Case "<<j<<":"<<endl;
j++;
cout<<x<<" + "<<y<<" = ";
for(i=maxlen-1;i>=0;i--)
cout<<sum[i];
cout<<endl;
if(t!=0)
cout<<endl;
memset(xx,0,1001*sizeof(int));//,把xx数组里的值清0，memset是按照字节进行赋值，一开始写成memset(xx,0,1001);错误，整型四个字节，需要1001*4，即1001*sizeof(int)
memset(yy,0,1001*sizeof(int));
memset(sum,0,1001*sizeof(int));
}
}
方法3：
#include <iostream>
#include <string.h>
using namespace std;
int main()
{
char x[1001],y[10001];
int xx[1001]={0},yy[1001]={0},sum[1001]={0},i,j=1,lenx,leny,maxlen;
int t;
cin>>t;
while(t--)
{
cin>>x>>y;
lenx=strlen(x);
leny=strlen(y);
maxlen=lenx>leny?lenx:leny;
for(i=0;i<lenx;i++)
xx[i]=x[lenx-i-1]-'0';
for(i=0;i<leny;i++)
yy[i]=y[leny-i-1]-'0';
for(i=0;i<maxlen;i++)
{
sum[i]+=xx[i]+yy[i];
if(sum[i]>=10)//判断两个大数的对应位相加是否大于等于10
{
sum[i+1]=1;//如果大于10，高位进1
sum[i]-=10;//如果大于20，本位-10
}
}
if(sum[i]==1)
maxlen++;
cout<<"Case "<<j<<":"<<endl;
j++;
cout<<x<<" + "<<y<<" = ";
for(i=maxlen-1;i>=0;i--)
cout<<sum[i];
cout<<endl;
if(t!=0)
cout<<endl;
memset(xx,0,1001*sizeof(int));
memset(yy,0,1001*sizeof(int));
memset(sum,0,1001*sizeof(int));
}
}
运行截图：
