Uva 424 - Integer Inquiry-优快云博客

本文详细阐述了一个使用高精度计算解决整数加法问题的实例，并深入讨论了在实现过程中遇到的问题及解决方案，包括数组大小设置、输出数组下标错误、进位计算等问题的解决方法。

Integer Inquiry

Time limit: 3.000 seconds

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
123456789012345678901234567890
123456789012345678901234567890
0

Sample Output

370370367037037036703703703670

#include<stdio.h>
#include<string.h>
int main()
{
    int sum[104], e, len, i, j, temp;
    char input[101];
    memset(sum, 0, sizeof(sum));
    memset(input, 0, sizeof(input));
    while(scanf("%s", input) == 1 && input[0])
    {
        getchar();
        e = 0;
        len = strlen(input);
        for(i=0; i<len; ++i)
        {
            temp = sum[i];
            sum[i] = (e + sum[i] + (input[len-i-1]-'0'))%10;
            e = (e + temp + (input[len-i-1]-'0'))/10;
        }
        for(j=i; j<104 && e; ++j)
        {
            temp = sum[j];
            sum[j] = (e + sum[j])%10;
            e = (e + temp)/10;
        }
        memset(input, 0, sizeof(input));
        
    }
    for(i=103, temp = 0; i>=0; --i)
    {
        if(sum[i] != 0)
        {
            temp = 1;
            printf("%d", sum[i]);
        }
        else if(temp == 1)
        {
            printf("%d", sum[i]);
        }
    }
    printf("\n");
    return 0;
}