houyunjie @ 2024-09-07 16:57:55
#include<bits/stdc++.h>
using namespace std;
string s1,s2;
int a[50000000],b[50000000],c[100000019];
int l;
int main(){
cin >> s1 >> s2;
for(int i = 0;i < s1.size();i++){
a[s1.size () - i - 1] = s1[i] - '0';
}
for(int i = 0;i < s2.size();i++){
b[s2.size () - i - 1] = s2[i] - '0';
}
l = s1.size();
if(s2.size() > l){
l = s2.size();
}
for(int i = 0;i < l;i++){
c[i]=a[i] + b[i];
}
for(int i = 0;i < l;i++){
if(c[i]>=10){
c[i + 1] = c[i + 1] + c[i] / 10;
c[i]%=10;
}
}
if(c[l] != 0){
l++;
}
for(int i = l - 1;i >= 0; i--){
cout<<c[i];
}
return 0;
}
by jijidawang @ 2024-09-07 17:01:28
有负数
by lxr_Galaxy @ 2024-09-07 17:06:31
@houyunjie az....这题不是高精直接加就行了
by houyunjie @ 2024-09-07 17:09:27
e..................................
by ZYStream @ 2024-09-07 22:19:34
666
by Chuex @ 2024-09-08 20:22:32
by Chuex @ 2024-09-08 20:26:04
敢问少侠是怎么用最难的思路做最简单的题而且还算错的。阁下难道就不能好好做题吗,有着思路留着AKIOI不好吗
by lkumy @ 2024-09-10 17:07:31
666
by litangzheng @ 2024-09-11 13:03:36
无敌了
by fengpengyu2030 @ 2024-09-11 17:47:44
@houyunjie 不用高精就行
by youyou_2025 @ 2024-09-16 16:27:14
@houyunjie 本题有负数,高精度过不去啊,你怎么宗名了