A + B Problem II(大数问题)
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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
21 2112233445566778899 998877665544332211
Sample Output
Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110
用字符串做:
#include<stdio.h>#include<string.h>char str1[1005],str2[1005],str[1005];int main(){ int t; while(scanf("%d",&t)!=EOF){ //scanf("%d",&t); getchar(); int cases=1; while(t--){ if(cases!=1) printf("\n"); scanf("%s%s",str1,str2); int i,j,temp=0,k=0; int len1=strlen(str1),len2=strlen(str2); for(i=len1-1,j=len2-1;i>=0 && j>=0;i--,j--){ temp+=(str1[i]-'0')+(str2[j]-'0'); str[k++]=(temp+'0'); temp/=10; } if(i<0 && j<0 && temp!=0) str[k++]=(temp+'0'); while(i>=0){ temp+=(str1[i--]-'0'); str[k++]=(temp+'0'); temp/=10; } while(j>=0){ temp+=(str2[j--]-'0'); str[k++]=temp+'0'; temp/=10; } if(temp!=0) str[k++]=(temp+'0'); printf("Case %d:\n",cases); printf("%s + %s = ",str1,str2); for(i=k-1;i>0;i--) printf("%c",str[i]); printf("%c\n",str[i]); cases++; } } return 0;}
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