NYOJ-A+B Problem II

A+B Problem II

时间限制：3000 ms  |  内存限制：65535 KB

难度：3

描述

I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

A,B must be positive.

输入

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.

输出

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.

样例输入

21 2112233445566778899 998877665544332211

样例输出

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110

代码:

#include<stdio.h>#include<string.h>char a[1010],b[1010];int s1[1010],s2[1010];int sum[1010];int main(){int T,k=0,i,j,d,c;scanf("%d",&T);while(T--){memset(s1,0,sizeof(s1));memset(s2,0,sizeof(s2));memset(sum,0,sizeof(sum));k++;scanf("%s %s",a,b);int len1,len2;len1=strlen(a);len2=strlen(b);for(i=len1-1,j=0;i>=0;--i,++j)s1[j]=a[i]-'0';for(i=len2-1,j=0;i>=0;--i,++j)s2[j]=b[i]-'0';c=0;for(i=0;i<len1||i<len2;++i){d=s1[i]+s2[i]+c;sum[i]=d%10;c=d/10;}printf("Case %d:\n",k);printf("%s + %s = ",a,b);if(c)sum[i++]=c;for(;i>0;--i)if(sum[i])break;for(;i>=0;--i)printf("%d",sum[i]);printf("\n");}return 0;}