A + B Problem II
Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 261413Accepted Submission(s): 50581
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
Author
Ignatius.L
思路：
将数字以字符的形式存储到字符数组中，因为在存储的时候是高位在以0为下标的下标变量中存储的，所以要将其进行翻转，存储到整形数组中（也就是高位存储到大下标变量中，因为在进位的时候能在原来的基础上进行i++，来存储最高位的数据），然后将两个大数按位相加，，如果比十大，进行进位操作！
代码：#include <stdio.h>#include <string.h>#define N 10005char a[N],b[N];int c[N],d[N];int main(){int n,i,j,k,len1,len2;scanf("%d",&n);k=n;while(n–){memset(c,0,sizeof(c));//每次都得清零，所以得放到while循环里面！memset(d,0,sizeof(d));getchar();scanf("%s%s",a,b);//空格也是scanf的分割符！len1=strlen(a);len2=strlen(b);for(i=len1-1,j=0;i>=0;i–)//因为需要逆序保存，所以应该设变量j从0开始！c[j++]=a[i]-'0';for(i=len2-1,j=0;i>=0;i–)d[j++]=b[i]-'0';for(i=0;i<1001;i++){c[i]+=d[i];if(c[i]>=10){c[i]-=10;c[i+1]++;}}printf("Case %d:\n%s + %s = ",k-n,a,b);for(i=1000;i>=0&&c[i]==0;i–);if(i>=0)for(;i>=0;i–){printf("%d",c[i]);}elseprintf("0");printf("\n");if(n!=0)printf("\n");}return 0;}
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