10013 - Super long sums
Time limit: 3.000 seconds
The Problem
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.
The Input
The first line of a input file is an integer N, then a blank line followed by N input blocks. The first line of an each input block contains a single number M (1<=M<=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 M 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 M.
There is a blank line between input blocks.
The Output
There is a blank line between output blocks.
Sample Input
240 44 26 83 733 07 92 8
Sample Output
4750470
完整代码:
/*0.915s*/
#include<cstdio>
int a[1000010], b[1000010], c[1000010];
int main()
{
int t, M, i, k, carry;
scanf("%d", &t);
while (t--)
{
scanf("%d", &M);
for (i = 0; i < M; ++i)
scanf("%d%d", &a[i], &b[i]);
carry = 0;
for (i = M - 1; i >= 0; --i)
{
k = carry + a[i] + b[i];
carry = k > 9 ? 1 : 0;
c[i] = k % 10;
}
for (i = 0; i < M; i++)
printf("%d", c[i]);
putchar(10);
if(t) putchar(10);
}
return 0;
}
超级长整数求和

本文介绍了一种解决超级长整数求和问题的方法。针对每个整数可能达到100万位的情况,通过逐位相加并进位的方式实现了两个大整数的加法运算。该方法适用于当标准整数类型无法满足存储需求时的大数运算。
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