题目描述
blablabla
样例
blablabla
算法1
(暴力枚举) $O(n^2)$
blablabla
时间复杂度
参考文献
C++ 代码
blablabla
算法2
(暴力枚举) $O(n^2)$
blablabla
时间复杂度
参考文献
C++ 代码
#include <stdio.h>
#include <string.h>
#define maxv 10000000
#define max(x,y) ((x) > (y)?(x):(y))
char a[maxv], b[maxv], c[maxv];
int main()
{
int d;
scanf("%s %s", a, b);
int len_a = strlen(a);
int len_b = strlen(b);
for (int i = len_a; i < maxv;i++)a[i] = '0';
for (int i = len_b; i < maxv; i++)b[i] = '0';
int len = max(len_a, len_b);
for (int i = 0, j = len_a - 1; i < j; i++, j--)a[i] = a[i] + a[j], a[j] = a[i] - a[j], a[i] = a[i] - a[j];
for (int i = 0, j = len_b - 1; i < j; i++, j--)b[i] = b[i] + b[j], b[j] = b[i] - b[j], b[i] = b[i] - b[j];
for (int i = 0; i <= len; i++) {
if (!i)c[i] = (a[i] + b[i] - 96) % 10 + 48, d = (a[i] + b[i] - 96) / 10;
else {
c[i] = (a[i] + b[i] - 96 + d) % 10 + 48;
if (a[i] + b[i] - 96 + d >= 10)d = (a[i] + b[i] - 96 + d) / 10;
else d = 0;
}
}
if (c[len] - 48)for (int i = len; i >= 0; i--)printf("%c", c[i]);
else for (int i = len - 1; i >= 0; i--)printf("%c",c[i]);
return 0;
}