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HDU-1002 A + B Problem II
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
经典的大数加法题目。
注意将数字字符转为数字是 减去‘0’ , 将数字转为数字字符是 加上‘0’。
#include <iostream> using namespace std; int t, tt, l; int main(void) { char numa[1001], numb[1001]; void bigsum(char [], char []); while(cin >> t) { tt = 0; l = t; while(t--) { if(tt <= l) tt++; cin >> numa >> numb; bigsum(numa, numb); } } return 0; } void bigsum(char a[], char b[]) { int sum[2000] = {0}, k = 0; int len_a = (int)strlen(a); int len_b = (int)strlen(b); for(int i = len_a-1, j = len_b-1; ; i--, j--) { if(i < 0 && j >= 0) { sum[k++] += (b[j]-‘0‘); } else if(i >= 0 && j < 0) { sum[k++] += (a[i]-‘0‘); } else if(i < 0 && j < 0) { break; } else { sum[k] += (a[i]-‘0‘) + (b[j]-‘0‘); if(sum[k++] >= 10) { sum[k-1] -= 10; sum[k] = 1; } } } k--; cout << "Case " << tt << ":" << endl; cout << a << " + " << b << " = "; for( ; k >= 0; k--) { cout << sum[k]; } cout << endl; if(tt != l) cout << endl; }
HDU-1002 A + B Problem II