NYOJ 103 A+B Problem II

A+B Problem II

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

2
1 2
112233445566778899 998877665544332211

样例输出

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

AC码：

#include
       
        
#include
        
          #include
         
           int main() { int T,i,k,len[2]; char str[1010],ch[1010]; int num[2][1010],max,count=0; scanf("%d",&T); while(T--) { scanf("%s",str); scanf("%s",ch); len[0]=strlen(str); len[1]=strlen(ch); max=len[0]>len[1] len[0]:len[1]; memset(num,0,sizeof(num)); for(i=0;str[i]!='\0';i++) { num[0][len[0]]=str[i]-'0'; len[0]--; } for(i=0;ch[i]!='\0';i++) { num[1][len[1]]=ch[i]-'0'; len[1]--; } for(i=1;i<=max;i++) { num[0][i]=num[0][i]+num[1][i]; if(num[0][i]>=10) { num[0][i+1]+=1; num[0][i]=num[0][i]%10; } } printf("Case %d:\n",++count); printf("%s + %s = ",str,ch); if(num[0][max+1]!=0) printf("%d",num[0][max+1]); for(i=max;i>0;i--) printf("%d",num[0][i]); printf("\n"); } return 0; }
         
        
       

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