标签：roc hdu while desc cas osi 加法 include void
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;
}
标签：roc hdu while desc cas osi 加法 include void
原文地址：http://www.cnblogs.com/limyel/p/6613714.html
