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HPU 1002 A + B Problem II【大数】
A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 261413 Accepted 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，由于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。由于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。假设比十大，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的切割符！ len1=strlen(a); len2=strlen(b); for(i=len1-1,j=0;i>=0;i--)//由于须要逆序保存，所以应该设变量j从0開始！ 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]); } else printf("0"); printf("\n"); if(n!=0) printf("\n"); } return 0;}
HPU 1002 A + B Problem II【大数】
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