简单高精加

#include <bits/stdc++.h>
using namespace std;

int main(){
    string a,b,c;
    bool f=0;
    cin >> a >> b;
    while(a.size()<b.size())a='0'+a;
    while(a.size()>b.size())b='0'+b;
    for(int i = a.size()-1;i>=0;i--){
        // cout << a[i] << " " << b[i] << "\n";
        if(f==0){
            if(a[i]+b[i]-'0'-'0'>9){
                f=1;
                c=char(a[i]+b[i]-'0'-10)+c;
            }
            else{
                c=char(a[i]+b[i]-'0')+c;
                f=0;
            }
        }
        else{
            if(a[i]+b[i]-'0'-'0'+1>9){
                f=1;
                c=char(a[i]+b[i]-'0'-9)+c;
            }
            else{
                c=char(a[i]+b[i]-'0'+1)+c;
                f=0;
            }
        }
    }
    if(f)cout << 1;
    cout << c;
    return 0;
}

Ex.A+B Problem 官方题解

思路

本题是 __int128 模板题，直接上代码：

代码

#include<bits/stdc++.h>
#define int long long
using namespace std;
__int128 read()
{
    __int128 res=0;
    char scan[1005];
    scanf("%s",scan);
    for(int i=0;i<(int)strlen(scan);i++)
        res*=10,res+=scan[i]-'0';
    return res;
}
void print(__int128 sum)
{
    if(sum>9)
        print(sum/10);
    putchar(sum%10+'0');
}
signed main()
{
    __int128 a=read(),b=read();
    print(a+b);
    return 0;
}

read 是输入 __int128 类型的函数，print 是输出。这里运用了递归。

简单高精加（

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
string jia(string a , string b){
    string c = "";
    int x[1005] = {} , y[1005] = {} , z[1005] = {};
    for(int i = a.size() - 1, j = 1; i >= 0 ; i-- , j++)x[j] = a[i] - '0';
    for(int i = b.size() - 1, j = 1; i >= 0 ; i-- , j++)y[j] = b[i] - '0';
    int len = max(a.size(), b.size());
    for(int i = 1; i <= len; i++){
        z[i] += x[i] + y[i];
        z[i + 1] += z[i] / 10; 
        z[i] %= 10;
    }
    if(z[len + 1] > 0)len++;
    for(int i = len; i >= 1 ; i--)c += char(z[i] + '0');
    return c;
}
string a , b;
int main(){
    cin >> a >> b;
    cout << jia(a , b);
    return 0;
}