九度OJ 1119：Integer Inquiry（整数相加） （大数运算）
- 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
- 提示：
-
注意输入数据中，VeryLongInteger 可能有前导0
- 来源：
- 2008年北京大学图形实验室计算机研究生机试真题
思路：
大数运算类题目，用字符串或数组来做。
代码：
#include <stdio.h>
#include <string.h>
#define M 105
void print(int *a)
{
int i, j;
for (i=0; i<M; i++)
{
if (a[i] != 0)
break;
}
if (i == M)
printf("0\n");
else
{
for (j=i; j<M; j++)
printf("%d", a[j]);
printf("\n");
}
}
int main(void)
{
int n, i, j;
int a[M], b[M];
char s[M];
memset(a, 0, M*sizeof(int));
while (scanf("%s", s) != EOF)
{
if (strcmp(s, "0") == 0)
break;
memset(b, 0, M*sizeof(int));
n = strlen(s);
j = M-n;
for (i=0; i<n; i++)
b[j++] = s[i]-'0';
for (i=M-1; i>0; i--)
{
a[i] += b[i];
if (a[i] >= 10)
{
a[i-1] ++;
a[i] -= 10;
}
}
}
print(a);
return 0;
}
/**************************************************************
Problem: 1119
User: liangrx06
Language: C
Result: Accepted
Time:0 ms
Memory:912 kb
****************************************************************/