九度OJ 1119：Integer Inquiry（整数相加）（大数运算）

最新推荐文章于 2022-01-16 10:02:22 发布

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最新推荐文章于 2022-01-16 10:02:22 发布

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原文链接：http://www.cnblogs.com/liangrx06/p/5083913.html

本博客介绍如何使用C语言处理大数求和问题，包括输入解析、大数相加以及输出处理等关键步骤。

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：679

解决：357

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
****************************************************************/

转载于:https://www.cnblogs.com/liangrx06/p/5083913.html