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.
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. Output a blank line between two test cases.
样例输入
2
1 2
112233445566778899 998877665544332211
样例输出
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
1 #include<stdio.h> 2 #include<string.h> 3 #include<ctype.h> 4 #include<math.h> 5 6 char a[1001],b[1001]; 7 8 void deal(char *s1, char *s2) 9 { 10 int len1,len2,i,t,flag; 11 char *p,*q; 12 int c[1003]={0}; 13 len1=strlen(s1); len2=strlen(s2); 14 i=t=flag=0; 15 printf("%s + ",a); 16 printf("%s = ",b); 17 p=strrev(s1); 18 q=strrev(s2); 19 while(*p || *q){ 20 if(*p) t+=*p-'0'; 21 if(*q) t+=*q-'0'; 22 t+=flag; 23 if(t>=10){t=t%10; c[i++]=t; flag=1;} 24 else {c[i++]=t; flag=0;} 25 p++; q++; 26 t=0; 27 } 28 if(flag) c[i]+=1; 29 else i--; 30 31 for(; i>=0; i--){ 32 printf("%d",c[i]); 33 } 34 } 35 36 void solve() 37 { 38 int i,n,t; 39 i=1; 40 scanf("%d",&n); 41 t=n; 42 getchar(); 43 while(n--){ 44 scanf("%s",a); scanf("%s",b); 45 printf("Case %d:\n",i); 46 deal(a,b); 47 if(i!=t) printf("\n\n"); 48 else printf("\n"); 49 i++; 50 memset(a,'\0',strlen(a)*sizeof(char)); 51 memset(b,'\0',strlen(b)*sizeof(char)); 52 } 53 } 54 55 int main() 56 { 57 solve(); 58 return 0; 59 }