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
-
2 1 2 112233445566778899 998877665544332211
- 样例输出
-
Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110
1 2 /*function: 减法 */ 3 /* date: 2012 0725 */ 4 /*name: 赵杰迪 */ 5 /*environment:Dev-c++ 4.9.9.2 */ 6 #include<stdio.h> 7 #include<string.h> 8 #define maxn 1010 9 int shu1[maxn],shu2[maxn]; 10 char str1[maxn],str2[maxn]; 11 int main() 12 { 13 int n,i,j,len1,len2,k; 14 scanf("%d",&n); 15 getchar(); 16 for(k=1;k<=n;k++) 17 { 18 scanf("%s",str1); 19 scanf("%s",str2); 20 printf("Case %d:\n%s + %s = ",k,str1,str2); 21 memset(shu1,0,sizeof(shu1)); 22 memset(shu2,0,sizeof(shu2)); 23 len1=strlen(str1); 24 len2=strlen(str2); 25 //printf("lenstr1=%d,lenstr2=%d",len1,len2);// 26 for(j=0,i=len1-1;i>=0;i--) 27 shu1[j++]=str1[i]-'0'; 28 for(j=0,i=len2-1;i>=0;i--) 29 shu2[j++]=str2[i]-'0'; 30 for(i=0;i<maxn;i++) 31 { 32 shu1[i]+=shu2[i]; 33 if(shu1[i]>=10) 34 { 35 shu1[i]%=10; 36 shu1[i+1]++; 37 } 38 } 39 //for(i=0;i<maxn;i++) 40 //printf("%d",shu1[i]);// 41 for(j=maxn-1;j>=0;j--) 42 if(shu1[j])break; 43 if(j==-1) 44 printf("0"); 45 else 46 { 47 for(i=j;i>=0;i--) 48 printf("%d",shu1[i]); 49 } 50 51 printf("\n"); 52 } 53 return 0; 54 } 55 56 57