ZOJ 1292 Integer Inquiry-优快云博客

本文介绍了一种处理非常大的整数相加的算法实现，使用C语言编程，并通过一个具体的例子来展示如何读取、处理和输出这些大整数的求和结果。

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

123456789012345678901234567890
123456789012345678901234567890
123456789012345678901234567890
0

Sample Output

370370367037037036703703703670

=====================================

可能会有童鞋卷我说这么个高精加的题放在这是不是水他们……好吧，就让我作为一个sample放着好了。毕竟，高精加中有好多需要注意的问题，尤其是在进位上。

Code
#include<stdio.h>
#define MAX 200

int strlen(char str[])
{
    int i = 0;
    while(str[i]) i++;
    return i;
}

int a[MAX], sum[MAX], length, span;
char str[MAX];

void clean(int a[])
{
    int i;
    for (i=0; i<MAX; i++) a[i] = 0;
}

void readin()
{
    int i;
    scanf("%s", str); length = strlen(str);
    for (i=0; i<length; i++) a[i] = str[length - i - 1] - '0';
}

void add()
{
    int i;
    span = span>length? span : length;

    for (i=0; i<span; i++) sum[i] += a[i];
    i = 0;
    while (sum[i]||i<span)
    {
        sum[i+1] += sum[i] / 10; sum[i] = sum[i] % 10;
        i++;
    }
    span = i;
}

void output()
{
    int i;
    for (i=span-1; i>=0; i--) printf("%d", sum[i]);
}

main()
{
    int n;
    scanf("%d", &n);

    while (n--)
    {
        readin();
        clean(sum); span = 0;

        while(!(a[0]==0&&length==1))
        {
            add();
            readin();
        }
        output();
        if (n) printf("\n\n");
    }

    return 0;
}

转载于:https://www.cnblogs.com/menie/archive/2009/06/10/1500776.html