hdu 1003 A + B Problem II 使用整型數組輕鬆實現大數求和

爲題如下：

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

 

 

 

解體代碼如下，實現很簡單，這裏就不說明實現過程

 

 

/**********************************************
    > File Name: 1002.c
    > Author: sea
    > Mail: [email protected]
    > Created Time: Mon 06 May 2013 02:01:19 PM CST
 **********************************************/

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

#define N 1101

int main() {
    char str1[N],str2[N];
    int sum[N+1],i,j,m,n,t,s1,s2,k,flag=1;
    scanf("%d",&i);
    while(i--) {
        k=N-1;
        memset(str1,0,sizeof(str1));
        memset(str2,0,sizeof(str2));
        memset(sum,0,sizeof(sum));
        scanf("%s %s",str1,str2);
        m=strlen(str1)-1;
        n=strlen(str2)-1;
        if(m>n) t=m;
        else t=n;
        for(j=0;j<=t;j++,m--,n--,k--) {
            if(m<0) s1=0;
            else s1=str1[m]-48;
            if(n<0) s2=0;
            else s2=str2[n]-48;
            if(sum[k]+s1+s2>=10)
                sum[k-1]+=1;
            sum[k]=(sum[k]+s1+s2)%10;

        }
        sum[k]>0?k:k++;
        printf("Case %d:\n%s + %s = ",flag++,str1,str2);
        for(j=k;j<N;j++)
            printf("%d",sum[j]);
        if(i>=1)    printf("\n\n");
        else printf("\n");
    }
    return 0;
}