九度OJ 1119 Integer Inquiry

题目描述：

    One of the first users of BIT's new supercomputer was Chip Diller.

    He extended his exploration of powers of 3 to go from 0 to 333 and he explored taking various sums of those numbers. 

    "This supercomputer is great,'' remarked Chip. 

    "I only wish Timothy were here to see these results.''

    (Chip moved to a new apartment, once one became available on the third floor of the Lemon Sky apartments on Third Street.)

输入：

    The input will consist of at most 100 lines of text, each of which contains a single VeryLongInteger. Each VeryLongInteger will be 100 or fewer characters in length, and will only contain digits (no VeryLongInteger will be negative).

    The final input line will contain a single zero on a line by itself.
输出：

    Your program should output the sum of the VeryLongIntegers given in the input.
样例输入：

123456789012345678901234567890
123456789012345678901234567890
123456789012345678901234567890
0

样例输出：

370370367037037036703703703670

题目意思是实现多个大整数相加。一直认为九度OJ上的题目出的意思都不是很明确，或者说不严谨。

我刚开始以为几个整数因为存在前导0，所以长度都一样，后来发现理解错了。这里的长度是可以不一样的。

我的思路是，每次输入一个str,就把它加到result中，最后输出result的值。输出的时候要把前导0去掉。还有如果两个00相加，结果是0，要把这时的0输出出来。

还有一点要注意，在每次把str加到result中时，最高位如果大于9，要把它进到高位去，以防出现溢出。

#include <iostream>
#include <cstring>
using namespace std;

int result[101];

void add(string str){ //实现将str加到result中
int len = str.length();
int i = 0;
for(i=0;i<len;i++){
int temp = result[100-i] + str[len-1-i] - '0';
result[100-i] = temp % 10;
result[100-i-1] += temp / 10;
}
while(result[100-i] > 9){ //将最高位的多位数化为个位数，
int temp = result[100-i];
result[100-i] = temp % 10;
result[100-i-1] += temp / 10;
i++;
}
}

int main(){
string str;
memset(result,0,sizeof(result));
while(cin>>str){
if(str == "0")
break;
else{
add(str);
}
}
bool flag = true;
int count = 0;
for(int i=0;i<101;i++){
if(result[i] == 0 && flag){ //把前导0去掉
count ++;
continue;
}
else{
flag = false;
cout<<result[i];
}
}
if(count == 101)
cout<<0<<endl; //表示0和0相加
cout<<endl;
return 0;
}