為題如下：

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，這里就不說明實現過程

/**********************************************?

????>?File?Name:?1002.c?

????>?Author:?sea?

????>?Mail:?windorman@gmail.com?

????>?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;?

}?

轉載于:https://blog.51cto.com/windorman/1195021

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