[HDOJ1002] A + B Problem II 高数度加法运算

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K


[align=left]Problem Description[/align]
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

2
1 2
112233445566778899 998877665544332211

[align=left]Sample Output[/align]

Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
[code]//4823648 2011-10-25 22:51:21 Accepted 1002 0MS 612K 1151 B C++ ajioy
#include <iostream>
using namespace std;
#define MAXLEN 1000
int main(){
char s1[MAXLEN],s2[MAXLEN];
int cases = 1,times;
cin >> times;
while(times--){
cin >> s1 >> s2;
int i;
int len1 = strlen(s1);
int len2 = strlen(s2);
int len = len1 >= len2 ? len1 : len2;
int *t1 = new int[(len + 1) * sizeof(int)];
int *t2 = new int[(len + 1) * sizeof(int)];
for(i = 0; i < len1; ++i) //把字符串转成整型数组,并反向以便运算
t1[i] = s1[len1 - i - 1] - '0';

for(i = 0; i < len2; ++i)
t2[i] = s2[len2 - i - 1] - '0';

for(i = len1; i < len + 1; ++i) t1[i] = 0; //最高位补0 ,与大数保持一致
for(i = len2; i < len + 1; ++i) t2[i] = 0;

for(i = 0; i < len; ++i) //普通相加
t1[i] += t2[i];

while(t1[len - 1] == 0 && len > 1) len--;//清除左边为0 的元素
for(i = 0; i < len; ++i)
if(t1[i] >= 10){
t1[i + 1] = t1[i + 1] + t1[i] / 10; //进位处理
t1[i] %= 10;
}

if(t1[i] != 0) //如果两数相加值超过较大数的位数
len = i + 1; //长度加1
cout << "Case " << cases << ":" << endl;
cout << s1 << " + " << s2 << " = "; //注意，+,=两边都有空格
for(i = len - 1;i >=0 ;--i)
cout << t1[i];
cout << endl;
if(times != 0) //最后一行没有空格
cout << endl;
cases++;
}

return 0;
}

[/code]