A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 150756 Accepted Submission(s): 28485
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
2
1 2
112233445566778899 998877665544332211
Sample Output
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
求两个数之和 大数求和
采用字符串读入加数 在进行计算存入一个数组中
1 #include <stdio.h> 2 #include <string.h> 3 4 #define N 1005 5 6 char ch[N],ch1[N]; 7 8 void add(char ch3[],char ch4[]) 9 { 10 int i = 0; 11 int j = 0; 12 int a[N]; 13 memset(a,0,sizeof(a)); 14 int k = 0; 15 int sum = 0; 16 int t = 0; 17 18 int len_ch = strlen(ch3); 19 int len_ch1 = strlen(ch4)-1; 20 21 for(i = len_ch-1; i>= 0; i--) 22 { 23 sum = ch3[i]+ch4[len_ch1] - 2*'0' + t; 24 a[k++] = sum %10; 25 t = sum /10; 26 len_ch1--; 27 } 28 29 for(j = len_ch1; j >= 0; j--) 30 { 31 sum = ch4[j] +t -'0'; 32 a[k++] = sum % 10; 33 t = sum /10; 34 } 35 36 for(i = k-1; i >= 0; i--) 37 printf("%d",a[i]); 38 printf("\n"); 39 } 40 41 int main() 42 { 43 int t = 0; 44 scanf("%d",&t); 45 int w = 1; 46 while(t--) 47 { 48 scanf("%s%s",ch,ch1); 49 printf("Case %d:\n",w++); 50 printf("%s + %s = ",ch,ch1); 51 if(strlen(ch) > strlen(ch1)) 52 add(ch1,ch); 53 else 54 add(ch,ch1); 55 if(t) printf("\n"); 56 } 57 return 0; 58 }