1002 hdu (高精度)
2014-03-13 14:14
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A + B Problem II
TimeLimit: 2000/1000 MS(Java/Others) MemoryLimit: 65536/32768 K (Java/Others)Total Submission(s):108726 AcceptedSubmission(s): 20548
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 integerT(1<=T<=20) which means the number oftest cases. Then T lines follow, each line consists of two positiveintegers, A and B. Notice that the integers are very large, thatmeans you should not process them by using
32-bit integer. You mayassume 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 isthe an equation "A + B = Sum", Sum means the result of A + B. Notethere are some spaces int the equation. Output a blank line
betweentwo test cases.
Sample Input
2 1 2112233445566778899 998877665544332211
Sample Output
Case1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 =1111111111111111110
Author
Ignatius.L
#include <stdio.h> #include<string.h> int sum[1000]; void add (char a[] ,char b[]) { int i= strlen(a) - 1 ,j = strlen (b) - 1; int x= 0 ,s ,k = 0; while(i >= 0 && j>= 0) { s = a[i--]+ b[j--] - 96 + x ; sum[k++] =s % 10 ; x = s / 10; } if (j== -1) while(i >= 0) { s = a[i--] - 48 + x ; sum[k++] = s % 10 ; x = s / 10 ; } if (i== -1) while(j >= 0) { s = b[j--] - 48 + x ; sum[k++] = s % 10 ; x = s / 10 ; } if(x) sum[k]= x ; else k = k- 1 ; for (i= k ;i >= 0 ;i--) printf("%d",sum[i]); } int main () { int t,i; chara[1000] ,b[1000]; scanf("%d",&t); for (i= 1 ;i <= t ;i++) { memset(a , '0' ,sizeof(a)); memset(b , '0' ,sizeof(b)); scanf ("%s%s",a ,b); printf ("Case %d:\n%s + %s =",i ,a ,b); add (a ,b); printf ("\n"); if(i != t)printf("\n"); } return0; }
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