POJ2602 Superlong sums水题

Posted 2021-02-02 tigerisland45

Superlong sums
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 23914 Accepted: 7086

Description

The creators of a new programming language D++ have found out that whatever limit for SuperLongInt type they make, sometimes programmers need to operate even larger numbers. A limit of 1000 digits is so small... You have to find the sum of two numbers with maximal size of 1.000.000 digits.

Input

The first line of an input file contains a single number N (1<=N<=1000000) - the length of the integers (in order to make their lengths equal, some leading zeroes can be added). It is followed by these integers written in columns. That is, the next N lines contain two digits each, divided by a space. Each of the two given integers is not less than 1, and the length of their sum does not exceed N.

Output

Output file should contain exactly N digits in a single line representing the sum of these two integers.

Sample Input

4
0 4
4 2
6 8
3 7

Sample Output

4750
Hint

Huge input,scanf is recommended.

Source

Ural State University collegiate programming contest 2000

问题链接：POJ2602 Superlong sums
问题简述：（略）
问题分析：
????大数计算的水题，要看程序是否写得简洁。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C语言程序如下：

/* POJ2602 Superlong sums */

#include <stdio.h>

#define N 1000000
char a[N + 1];

int main(void)
{
    int n, b, carry, i;
    scanf("%d", &n);
    for(i = 0; i < n; i++) {
        scanf("%d%d", &a[i], &b);
        a[i] += b;
    }
    for(carry = 0, i--; i >= 0; i--) {
        a[i] += carry;
        carry = a[i] / 10;
        a[i] %= 10;
        a[i] += '0';
    }

    puts(a);

    return 0;
}