POJ – 1503 Integer Inquiry Integer Inquiry

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本文介绍了解决POJ 1503 IntegerInquiry问题的方法，该问题是关于大数加法的计算。文章提供了一种使用数组来存储和计算大数的解决方案。

POJ –1503 Integer Inquiry

Integer Inquiry

Time Limit: 1000MS	Memory Limit: 10000K
Total Submissions: 34728	Accepted: 13523

Description

One of the firstusers of BIT's new supercomputer was Chip Diller. He extended his explorationof powers of 3 to go from 0 to 333 and he explored taking various sums of thosenumbers.
``This supercomputer is great,'' remarked Chip. ``I only wish Timothy were hereto see these results.'' (Chip moved to a new apartment, once one becameavailable on the third floor of the Lemon Sky apartments on ThirdStreet.)

Input

The input willconsist 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 onlycontain digits (no VeryLongInteger will be negative).

The final input line will contain a single zero on a line by itself.

Output

Your programshould output the sum of the VeryLongIntegers given in the input.

Sample Input

123456789012345678901234567890

Sample Output

370370367037037036703703703670

大数加法----》直接用java中的BigInteger应该时最快的方法，不过我这里通过数组实现：

将大数的每位分配到数组中的每一位然后相加。

代码：

//poj 1503 Integer Inquiry
#include <cstdio>
#include <cstring>
const int max = 200;
int main(int argc,char const *argv[])
{
    //freopen("in.txt","r", stdin);
    int sum[max];
    memset(sum,0, sizeof sum);
    char line[150];
    //waring the gets
    while (gets(line)&& !(line[0]== '0'&&line[1]== '\0')){
        int s =0, c= 0;
        int n =0, j= max -1;
        while (line[n++]);
        n -= 2;
        while (n >= 0) {
            int tmp = line[n]- '0';
            s = sum[j]+ tmp + c;
            c = s/ 10;
            sum[j]= s %10;
            n--; j--;
        }
        while (c) {
            s = sum[j]+ c;
            c = s/ 10;
            sum[j--]= s %10;
        }
    }
    for (int i =0; i< max; i++){
        if (sum[i]){
            for (; i < max; i++)
                printf("%d", sum[i]);
        }
    }
    printf("\n");
    return 0;
}