A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 209837 Accepted Submission(s): 40373
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
Author
Ignatius.L
一样的题目，不一样的做法；
#include<stdio.h>
#include<string.h> #include<stdlib.h> #define M 1000 char a[M]; char b[M]; char sum1[M]; char sum2[M]; int sum[M+100]; int main() { int T,ca; scanf("%d",&T); for(ca=1;ca<=T;ca++) { memset(a,sizeof(a)); memset(b,sizeof(b)); memset(sum1,sizeof(sum1)); memset(sum2,sizeof(sum2)); memset(sum,sizeof(sum)); int i,j,l,d,t; scanf("%s%s",a,b); for(i=strlen(a)-1,j=0;i>=0;i--,j++) sum1[j]=a[i]-'0'; for(i=strlen(b)-1,j++) sum2[j]=b[i]-'0'; l=strlen(a)>=strlen(b)?strlen(a):strlen(b); for(i=0;i<l;i++) sum[i]=sum1[i]+sum2[i]; for(i=0;i<l;i++) { if(sum[l-1]>9) L++; if(sum[i]>9) { sum[i]-=10; sum[i+1]+=1; } } printf("Case %d:\n",ca); printf("%s + %s = ",b); for(i=l-1;i>=0;i--) { printf("%d",sum[i]); } printf("\n"); if(ca<T) printf("\n"); } }
