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A + B Problem II
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
#include<cstdio>#include<cstring>void plus(int *a,int *b){ int k; k=a[0]>b[0]?a[0]:b[0]; for(int i=1; i<=k; i++) { a[i+1]+=(a[i]+b[i])/10; a[i]=(a[i]+b[i])%10; } if(a[k+1]) a[0]=k+1; else a[0]=k;}int main(){ char s1[1005],s2[1005]; int a[1005],b[1005]; int T,i,n,m,t; scanf("%d",&T); for(t=1;t<=T;t++) { n=m=1; memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); scanf("%s",s1); scanf("%s",s2); for(i=strlen(s1)-1; i>=0; i--) a[n++]=s1[i]-‘0‘; a[0]=n-1; for(i=strlen(s2)-1; i>=0; i--) b[m++]=s2[i]-‘0‘; b[0]=m-1; printf("Case %d:\n",t); printf("%s + %s",s1,s2); printf(" = "); plus(a,b); for(i=a[0]; i>=1; i--) printf("%d",a[i]); printf("\n"); if(t!=T) printf("\n"); } return 0;}/*前几次一直PE，后来才发现原来是最后面多输出了一个回车，囧*/
A + B Problem II
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