Description
求N进制的高精度加法
Input
第一行输入N（2≤N≤10）
第二行两个数X Y（长度均≤100）
Output
输出N进制下X和Y的和
Sample Input
5
2 4
Sample Output
11
#include <cstdio>
#include <iostream>
#include <cmath>
#include <
string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
using namespace std;
#define ll long long
#define eps 1e-9
const int inf =
0x3f3f3f3f;
const int mod = 1e9+
7;
char ads[
10000005];
char s1[
10005], s2[
10005];
int n;
void additive(
char* a,
char*
b)
{
memset(ads, 0,
sizeof(ads));
int len1, len2;
len1 =
strlen(a);
len2 =
strlen(b);
if(len1 < len2)
///保持len1是最长的
{
for(
int i =
0; i<len2; i++
)
swap(a[i], b[i]);
swap(len1, len2);
}
reverse(a, a +
len1);
reverse(b, b +
len2);
int t =
0, i;
for(i =
0; i < len2; i++)
///一遍求和一边求进制
{
ads[i] = ((a[i] -
'0') + (b[i] -
'0') + t) % n +
'0';
t = (a[i] + b[i] -
'0' -
'0' + t) /
n;
}
for(
int k = i; k < len1; k++)
///处理较大的那个数没有加的那部分
{
ads[k] = (a[k] + t -
'0') % n +
'0';
t = (a[k] + t -
'0') /
n;
}
if(t !=
0)
printf("%d", t);
for(
int m = len1 -
1; m >=
0; m--
)
printf("%c", ads[m]);
printf("\n");
}
int main()
{
scanf("%d", &
n);
cin>>s1>>
s2;
additive(s1, s2);
return 0;
}
转载于:https://www.cnblogs.com/RootVount/p/11340424.html
