hduoj 1002 A + B Problem II
原題鏈接:http://acm.hdu.edu.cn/showproblem.php?pid=1002
題目描述如下:
A + B Problem II
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/32768 K (Java/Others)
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
題目大意: 大數相加
題目思路: 從最低位開始一位一位的加。註意進位,註意清除前綴的0
C++ 代碼如下:
#include <cstring> #include <iostream> using namespace std; int main() { int n; char a[1001],b[1001]; short c[1001]; cin >> n; for (int i=0; i<n; i++) { int alen,blen,clen,maxlen; cin >> a >> b; alen = strlen(a); blen = strlen(b); clen = 0; maxlen = alen/blen ? alen : blen; int ai,bi,s=0; for (int j=0; j<maxlen; j++) { ai = alen-j-1; bi = blen-j-1; if (ai>=0 && bi>=0) s = a[ai]+b[bi]-2*‘0‘+s/10; else if (ai>=0) s = a[ai]-‘0‘+s/10; else if (bi>=0) s = b[bi]-‘0‘+s/10; c[clen++] = s%10; } if (s/10) c[clen++] = s/10; for (int k=clen-1; k>=0; k--) if (c[k]) break; else clen--; cout << "Case " << i+1 << ":" << endl; cout << a <<" + " << b << " = "; for (int k=clen-1; k>=0; k--) cout << c[k]; cout << endl; if (n-i-1) cout << endl; } return 0; }
hduoj 1002 A + B Problem II