杭电OJ题 1047 Integer Inquiry 解题报告

最新推荐文章于 2019-03-24 15:25:45 发布

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本文详细介绍了杭电在线判题系统（OJ）中编号为1047的Integer Inquiry题目，包括问题背景、解题思路及具体解决方案。通过深入探讨大数据数的处理技巧，展示如何在ACM竞赛环境中应对BigNumber类问题。

Integer Inquiry

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

Total Submission(s): 8039 Accepted Submission(s): 2058

Problem Description

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.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Sample Input

1


123456789012345678901234567890
123456789012345678901234567890
123456789012345678901234567890
0

Sample Output

370370367037037036703703703670

————————————————————————————————————————————————

注意细节的处理

/****************************
 *Name:Integer Inquiry.c
 *Tags:ACM BigNumber
 *Note:注意细节
 ****************************/

#include <stdio.h>
#include <string.h>

int main()
{
      int N, num[101], i, j, c, l, min;
      char str[101];
      scanf("%d", &N);
      getchar();
      while(N--) {
	    getchar();
	    for(i = 0; i < 101; i++) {
		  num[i] = 0;
	    }
	    j = 100;
	    min = 101;
	    while(gets(str) && (str[1] != '\0' || str[0] != '0')) {
		  l = strlen(str);
		  for(i = l-1, j = 100; i >= 0; i--, j--) {
			num[j] += (str[i]-'0');
		  }
		  if(j < min) {
			min = j; //注意记录每次数据的最长大小
		  }
	    }
	    c = 0;
	    j = min;
	    for(i = 100; i > j; i--) {
		  num[i] += c;
		  c = num[i] / 10;
		  num[i] = num[i] % 10;
	    }
	    if(c) {
		  num[j] = c;
	    }
	    while(j < 101 && !num[j]) {
		  j++;
	    } //处理全为零的情况
	    if(j == 101) {
		  printf("0\n");
	    } else {
		  while(j < 101) {
			printf("%d", num[j]);
			j++;
		  }
		  printf("\n");
	    }
	    if(N) {
		  printf("\n");
	    }
      }
      return 0;
}