hdu 1002 A + B Problem II

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 150756    Accepted Submission(s): 28485

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

 

求两个数之和 大数求和

 

采用字符串读入加数 在进行计算存入一个数组中

 

#include <stdio.h>
#include <string.h>

#define N 1005

char ch[N],ch1[N];

void add(char ch3[],char ch4[])
{
    int i = 0;
    int j = 0;
    int a[N];
    memset(a,0,sizeof(a));
    int k = 0;
    int sum = 0;
    int t = 0;

    int len_ch = strlen(ch3);
    int len_ch1 = strlen(ch4)-1;

    for(i = len_ch-1; i>= 0; i--)
    {
        sum = ch3[i]+ch4[len_ch1] - 2*'0' + t;
        a[k++] = sum %10;
        t = sum /10;
        len_ch1--;
    }

    for(j = len_ch1; j >= 0; j--)
    {
        sum = ch4[j] +t -'0';
        a[k++] = sum % 10;
        t = sum /10;
    }

    for(i = k-1; i >= 0; i--)
        printf("%d",a[i]);
        printf("\n");
}

int main()
{
    int t = 0;
    scanf("%d",&t);
    int w = 1;
    while(t--)
    {
        scanf("%s%s",ch,ch1);
        printf("Case %d:\n",w++);
        printf("%s + %s = ",ch,ch1);
        if(strlen(ch) > strlen(ch1))
            add(ch1,ch);
        else
        add(ch,ch1);
        if(t) printf("\n");
    }
    return 0;
}