Hdu oj 1002 A + B Problem II（大数加法）

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 308383    Accepted Submission(s): 59612

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

21 2112233445566778899 998877665544332211

 

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110
 
大数加法，模拟手算，注意输出格式
#include<cstdio>#include<iostream>#include<cstring>using namespace std;const int maxn = 1005;char a[maxn],b[maxn];int c[maxn];int main(){    int t;    scanf("%d",&t);    int ans = 0;    while(t--)    {        getchar();        ans++;        scanf("%s%s",&a,&b);        int len_a = strlen(a);        int len_b = strlen(b);        int temp = 0;        int cnt = 0;        if(len_a >= len_b)        {            for(int i=len_b-1;i>=0;i--)            {                c[cnt] = ((a[len_a - cnt - 1] - '0') + (b[i] - '0') + temp) % 10;                temp = ((a[len_a - cnt - 1] - '0') + (b[i] - '0') + temp) / 10;                cnt++;            }            for(int i=len_a-len_b-1;i>=0;i--)            {                c[cnt] = ((a[i] - '0')+ temp) % 10;                temp = ((a[i] - '0') + temp) / 10;                cnt++;            }        }        else        {            for(int i=len_a-1;i>=0;i--)            {                c[cnt] = ((b[len_b - cnt - 1] - '0') + (a[i] - '0') + temp) % 10;                temp = ((b[len_b - cnt - 1] - '0') + (a[i] - '0') + temp) / 10;                cnt++;            }            for(int i=len_b-len_a-1;i>=0;i--)            {                c[cnt] = ((b[i] - '0')+ temp) % 10;                temp = ((b[i] - '0') + temp) / 10;                cnt++;            }        }        printf("Case %d:\n",ans);        printf("%s + %s = ",a,b);        if(temp)            printf("%d",temp);        for(int i=cnt-1;i>=0;i--)        {            printf("%d",c[i]);        }        printf("\n");        if(t)            printf("\n");    }}