HDU1002

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A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 275474 Accepted Submission(s): 53177

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

大数计算用字符串写的虽然是水题但是很坑的。。

#include <iostream>#include <cstdio>#include <string.h>#define B 1#define A 0using namespace std;int main(){    int ncase;    while(scanf("%d",&ncase)!=EOF)    {        int mark;        for(int m=1;m<=ncase;m++)        {            char a[1005];            char b[1005];            int  c[1005];            char tran[1005];            scanf("%s",a);            scanf("%s",b);            memset(c,0,sizeof(c));            if(m==1)            {                cout<<"Case "<<m<<":"<<endl;                cout<<a<<" + "<<b<<" = ";            }            else            {                cout<<endl;                cout<<"Case "<<m<<":"<<endl;                cout<<a<<" + "<<b<<" = ";            }            int lena=strlen(a);            int lenb=strlen(b);            int max,min;            if(lena>lenb)            {                int m=0;                for(int i=lena-lenb;i<=lena-1;i++)                {                    tran[i]=b[m];                    m++;                }                for(int i=0;i<=lena-lenb-1;i++)                {                    tran[i]='0';                }                memcpy(b,tran,sizeof(a));                max=lena;                min=lenb;            }            else            {                int m=0;                for(int i=lenb-lena;i<=lenb-1;i++)                {                    tran[i]=a[m];                    m++;                }                for(int i=0;i<=lenb-lena-1;i++)                {                    tran[i]='0';                }                memcpy(a,tran,sizeof(a));                max=lenb;                min=lena;            }            for(int i=max-1;i>0;i--)            {                c[i]=a[i]-'0'+b[i]-'0'+c[i];                if(c[i]>9)                {                    c[i-1]++;                    c[i]=c[i]-10;                }            }            if((a[0]-'0'+b[0]-'0'+c[0])>9)            {                cout<<"1";                c[0]=a[0]-'0'+b[0]-'0'-10+c[0];            }            else            {                c[0]=a[0]-'0'+b[0]-'0'+c[0];            }            for(int i=0;i<=max-1;i++)            {                cout<<c[i];            }            cout<<endl;        }    }}

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