A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 260025 Accepted Submission(s): 50264
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
Author
Ignatius.L
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思路:用数组处理大数问题,用p表示进位
注意格式:Case %d:\n %s + %s = (空格问题)
思路:用数组处理大数问题,用p表示进位
注意格式:Case %d:\n %s + %s = (空格问题)
#include<stdio.h> #include<string.h> int main(){ int n,alen,blen,i,j=1,k,p=0; char a[1000],b[1000],c[1001]; scanf("%d",&n); while(n){//不能是n-- scanf("%s %s",a,b); printf("Case %d:\n",j); printf("%s + %s = ",b); alen=strlen(a)-1; blen=strlen(b)-1; for(k=0;alen>=0||blen>=0;alen--,blen--,k++){ if(alen>=0&&blen>=0) c[k]=a[alen]+b[blen]-'0'+p; if(alen<0&&blen>=0) c[k]=b[blen]+p; if(alen>=0&&blen<0) c[k]=a[alen]+p; p=0; //进位加上之后要清零 if(c[k]>'9') { c[k]-=10; p=1; } } if(p==1) printf("1");//最后的进位p=1 while(k--) printf("%c",c[k]); j++; if(n!=1) printf("\n\n"); else printf("\n"); n--;//此处n-- } return 0; }