UVa 424 Integer Inquiry（整数查询）

Integer Inquiry

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

题目大意：

  大整数相加，实际上像是小学的竖式求和，一位一位的相加就可以了。

注意点：WA了两次，后来发现，原来最后输出之后，要有个换行符！！！

后来发现，不一定需要判断最后的一行是0，可以一直读到EOF。

下面贴出两种不同想法的代码：

代码一：

将所有行都读完，再来进行大数计算。

#include 
#include 
#include 
#define INF 1e9

char str[105][105];
int num[105][105];
int ans[10010];

int main(){
    int i, j;
    int len, count, max = -INF, flag;

    i = 0;
    while(~scanf("%s", str[i])){
        if(str[i][0] == '0')  break;
        i++;
    }

    count = i;
    memset(num, 0, sizeof(num));
    for(i = 0; i0; i--){
        if(flag){
            printf("%d", ans[i]);
        }
        else if(ans[i]){
            flag = 1;
            printf("%d", ans[i]);
        }
    }
    printf("\n");
    return 0;
}

代码二：

没读一行字符串，进行一次大数相加。

#include 
#include 
#include 
#include 

char str[100];
int num[100];
int ans[10000];

int main(){
    int i, j;
    int len, flag;

    memset(ans, 0, sizeof(ans));
    while(~scanf("%s", str)){
        len = strlen(str);
        memset(num, 0, sizeof(num));
        for(i = 0; i0; i--){
        if(flag)
            printf("%d", ans[i]);
        else if(ans[i]){
            printf("%d", ans[i]);
            flag = 1;
        }
    }
    printf("\n");
    return 0;
}