A + B Problem II

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A + B Problem IITime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 169117    Accepted Submission(s): 32432 Problem DescriptionI have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.  InputThe 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.  OutputFor 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 Input2 1 2 112233445566778899 998877665544332211 Sample OutputCase 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110  #include <iostream> #include <cstring> using namespace std; void add(char a[],char b[]) { char sum[1010]={' '}; int flg=0; int temp =0; int len_a =strlen(a); int len_b =strlen(b); int i=len_a; int j=len_b; for (;i>0;i--) { if (j>0) { temp =a[i-1]+b[j-1]+flg-96; j--; } else temp = a[i-1]+flg-48; if (temp>=10) { flg=1; } else flg =0; temp =temp%10; sum[i]=temp+48; } if (flg==1)sum[0]=49; i=0; while (i<=len_a) { if (sum[i]!=' ')cout<<sum[i]; i++; } cout<<endl; } int main() {int N; while (cin >>N ) { for (int i=1;i<=N;i++) { char a[1000]; char b[1000]; cin >>a; cin >>b; int len_a =strlen(a); int len_b =strlen(b); cout <<"Case "<<i<<":\n"<<a<<" + "<<b<<" = "; if (len_a>=len_b) { add(a,b); } else add(b,a); if (i!=N)cout<<endl; } } return 0; }

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