http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=365
424 - Integer Inquiry
Time limit: 3.000 seconds
| Integer Inquiry |
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.)
Input
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.
Output
Your program should output the sum of the VeryLongIntegers given in the input.
Sample Input
123456789012345678901234567890 123456789012345678901234 567890 123456789012345678901234 567890 0
Sample Output
370370367037037036703703703670
参考http://www.voidcn.com/article/p-ydvykbtz-bce.html的发现可以用白书73页的模版做。。原来一直不知道73页怎么用，原来是这样的。。谢谢最后的午餐了。。
#include <cstdio>
#include <iostream>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
#define MAXN 80 + 10
#define MAX 255 + 10
const int maxn = 1000;
int max(int a,int b)
{
return a>b?a:b;
}
// 大数类的定义
struct bign
{
int len,s[maxn];
bign()
{
memset(s,sizeof(s));
len = 1;
}
// 重载操作符‘=’用于赋值字符数组
bign operator = (const char* num)
{
len = strlen(num);
for(int i = 0; i < len; i++)
{
s[i] = num[len - i - 1] - '0';
}
return *this;
}
// 重载操作符‘=’用于赋值数字
bign operator = (int num)
{
char s[maxn];
sprintf(s,"%d",num);
*this = s;
return *this;
}
// 参数为数字类型的构造函数
bign(int num)
{
*this = num;
}
// 参数为字符数组类型的构造函数
bign(const char* num)
{
*this = num;
}
// 将字符数组转化成字符串
string