大数问题（相加） A + B

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

AC代码：

#include<iostream>
#include<string.h>
using namespace std;
#include<stdio.h>
#define MAX 1010
int add1[MAX], add2[MAX], res[MAX];
char tmp1[MAX], tmp2[MAX];
int main()
{
    int N, i, j, len, len1, len2, tmp, k;
        scanf("%d",&N);
        getchar();
        for(j=;j<N;j++)
        {
            memset(add1,,sizeof(add1));
            memset(add2,,sizeof(add2));
            memset(res,,sizeof(res));
            memset(tmp1,,sizeof(tmp1));
            memset(tmp2,,sizeof(tmp2));
            scanf("%s %s",tmp1,tmp2);
            len1 = strlen(tmp1);
            len2 = strlen(tmp2);
            for(i=len1-,k=;i>=;i--)
                add1[k++] = tmp1[i] - '0';

            for(i=len2-,k=;i>=;i--)
                add2[k++] = tmp2[i] - '0';
            tmp = ;
            if(len1 >= len2)
            {
                for(i=;i<=len1;i++)
                {
                    res[i] = (add1[i] + add2[i] +tmp)%;
                    tmp = (add1[i] + add2[i] +tmp)/;
                }
            }
            else if(len1 < len2)
            {
                for(i=;i<=len2;i++)
                {
                    res[i] = (add1[i] + add2[i] +tmp)%;
                    tmp = (add1[i] + add2[i] +tmp)/;
                }
            }
            if(len1 >= len2) len = len1;
            else len = len2;
            printf("Case %d:\n%s + %s = ",j+, tmp1 , tmp2);
            if(res[len]!=) printf("%d",res[len]);
            for(i=len-;i>=;i--)
                    printf("%d",res[i]);

            printf("\n");
            if(j!=N-) printf("\n");
        }
    return ;
}

my AC:

#include<iostream>
#include<stdio.h>
#include<string.h>
#define N 1010
using namespace std;
char a[N],b[N];
int c[N],d[N],ans[N];

int main(){
    int t ;
    cin>>t;
    getchar();
    for(int i=;i<=t;i++)
    {
        memset(a,,sizeof(a));
        memset(b,,sizeof(b));
        memset(c,,sizeof(c));
        memset(d,,sizeof(d));
        memset(ans,,sizeof(ans));
        cin>>a>>b;
        int la=strlen(a);
        int lb=strlen(b);
        for(int i=;i<la;i++) c[i]=a[la--i]-'';
        for(int i=;i<lb;i++) d[i]=b[lb--i]-'';
        int lans;
        (la>lb)?lans=la:lans=lb;
        int temp=;
        for(int i=;i<=lans;i++){
            ans[i]=(c[i]+d[i]+temp)%;
            temp=(c[i]+d[i]+temp)/;
        }
        cout<<"Case "<<i<<":"<<endl;
        cout<<a<<" + "<<b<<" = ";
        for(int i=lans-;i>=;i--)
        cout<<ans[i];
        cout<<endl;
        if(i!=t)cout<<endl;
    }

}

#include<iostream>
#include<stdio.h>
#include<string.h>
#define N 1010
using namespace std;
char a[N],b[N];
int c[N],d[N],ans[N];

int main(){
    int t ;
    cin>>t;
    getchar();
    for(int i=;i<=t;i++)
    {
        fill(a,a+N,);
        fill(b,b+N,);
        fill(c,c+N,);
        fill(d,d+N,);
        fill(ans,ans+N,);
        cin>>a>>b;
        int la=strlen(a);
        int lb=strlen(b);
        for(int i=;i<la;i++) c[i]=a[la--i]-'';
        for(int i=;i<lb;i++) d[i]=b[lb--i]-'';
        int lans;
        (la>lb)?lans=la:lans=lb;
        int temp=;
        for(int i=;i<=lans;i++){
            ans[i]=(c[i]+d[i]+temp)%;
            temp=(c[i]+d[i]+temp)/;
        }
        cout<<"Case "<<i<<":"<<endl;
        cout<<a<<" + "<<b<<" = ";
        for(int i=lans-;i>=;i--)
        cout<<ans[i];
        cout<<endl;
        if(i!=t)cout<<endl;
    }

}

#include<iostream>
#include<string.h>
using namespace std;
#include<stdio.h>
#define MAX 1010
int add1[MAX], add2[MAX], res[MAX];
char tmp1[MAX], tmp2[MAX];
int main()
{
int N, i, j, len, len1, len2, tmp, k;
scanf("%d",&N);
getchar();
for(j=;j<N;j++)
{
memset(add1,,sizeof(add1)) ;    ||用0填充
memset(add2,,sizeof(add2));
memset(res,,sizeof(res));
memset(tmp1,,sizeof(tmp1));
memset(tmp2,,sizeof(tmp2));

tips：函数解释

void *memset(void *s, int ch, size_t n);

函数解释：将s中前n个字节替换为ch并返回s；

cin>>tmp1>>tmp2;

量 数 组 长 度：
len1 = strlen(tmp1);
len2 = strlen(tmp2);
for(i=len1-,k=;i>=;i--)||改类型并换顺序
add1[k++] = tmp1[i] - '';

for(i=len2-,k=;i>=;i--)
add2[k++] = tmp2[i] - '';
tmp = ;
if(len1 >= len2)
{
for(i=;i<=len1;i++)
{
res[i] = (add1[i] + add2[i] +tmp)%;
tmp = (add1[i] + add2[i] +tmp)/;
}
}
else if(len1 < len2)
{
for(i=;i<=len2;i++)
{

进位处理方法：
res[i] = (add1[i] + add2[i] +tmp)%;
tmp = (add1[i] + add2[i] +tmp)/;
}
}
if(len1 >= len2) len = len1;
else len = len2;
printf("Case %d:\n%s + %s = ",j+, tmp1 , tmp2);
if(res[len]!=) printf("%d",res[len]);
for(i=len-;i>=;i--)
printf("%d",res[i]);

printf("\n");
if(j!=N-) printf("\n");
}
return ;
}

char tmp1[MAX], tmp2[MAX];

cin>>tmp1>>tmp2;

一长串的字符数组，还可以直接用cin进行输入

2. getchar（）必须加

3.字符型与整数类型互换

for(i=len1-1,k=0;i>=0;i--)
add1[k++] = tmp1[i] - '0';

for(i=len2-1,k=0;i>=0;i--)
add2[k++] = tmp2[i] - '0';

其实呢-48也是一样的；

4.错误代码

#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
int main()
{
int t;
long sa,sb;
cin>>t;
getchar();
for(int i=;i<=t;i++)
{
char a[],b[];
for(int j=;;j++)
{
char word=getchar();
if(word>=''&&word<='')
a[j]=word;
else
{
sa=j;
break;
}
}
for(int k=;;k++)
{
char word=getchar();
if(word>=''&&word<='')
b[k]=word;
else if(word=='\n')
{
sb=k;
break;
}
}
cout<<"Case "<<i<<":"<<endl;
for(int m=;m<sa;m++)
{
cout<<a[m];
}
cout<<" + ";
for(int n=;n<sb;n++)
{
cout<<b[n];
}
cout<<" = ";

int c[]={};编译有警告
int sc,next,shu,shua,shub;
if(sa>sb)
sc=sa;
else
sc=sb;       出错数据举例， ，所给出结果数组的长度设置有问题，应该分情况讨论，有0单独分情况，并设置不显示
for(int o=sc-;o>=;o--,sa--,sb--)
{
if(sa>)
shua=a[sa-]-;
else
shua=if(sb>)

shub=b[sb-]-;
else
shub=;
shu=shua+shub;
if(shu>=)
{
c[o-]=;
shu=shu-;
c[o]+=shu;       进位不全面， ，数据出错，所以不适宜顺序相加机制，应该采用逆序相加，顺序输出的方式
}
else
}

for(int p=;p<sc;p++)
cout<<c[p];
cout<<endl;
if(i!=t)
cout<<endl;
}
#include<iostream>