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>

巴特西

大数问题（相加） A + B

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