40分！高精度加法！大佬求助！

yinjunjie2012 @ 2024-08-23 09:35:02

#include<bits/stdc++.h>
using namespace std;
int main(){
    char a[100010],b[100010];
    cin>>a>>b;
    int a1[100010],b1[100010],la=strlen(a),lb=strlen(b);
    for(int i=0;i<la;i++)a1[i]=a[la-i-1]-48;
    for(int i=0;i<lb;i++)b1[i]=b[lb-i-1]-48;
    int c1[100010],lc=max(la,lb)+1;
    int x=0;
    for(int i=0;i<lc;i++){
        c1[i]=a1[i]+b1[i]+x;
        x=c1[i]/10;
        c1[i]%=10;
    }
    int flag=0;
    for(int i=lc;i>=0;i--){
        if(flag==0 and c1[i]==0){
            continue;
        }
        if(flag==0 and c1[i]!=0){
            cout<<c1[i];
            flag=1;
        }
        else{
            cout<<c1[i];
        }
    }
    return 0;
}

by jiangyixuan_eason @ 2024-08-28 12:49:11

这样不就行了吗，其实完全不用那么复杂呀

#include<bits/stdc++.h>
using namespace std;
int main(){
    int a,b;
    cin >>a>>b;
    cout <<a+b<<endl;
    return 0;
}

by guimei121212 @ 2024-08-30 18:52:17

@yinjunjie2012 像你这样用高精度做P1601双倍经验哦

by AAAbbb12321 @ 2024-09-06 20:02:05

额

by Liuchuhao @ 2024-10-20 16:20:20

print(int(inpit())+int(input()))

使用python,无惧高精度
违规紫杉

by Lucashuang @ 2024-11-03 13:19:44

二分+队列+递归更简单捏

#include<bits/stdc++.h>
using namespace std;
int m , n , a[200001] , l1 , s , r1 , maxx = 0 , ans;
priority_queue<int> que;
int fen(int l , int r){
    if(l == r) return a[l];
    m = (l + r) / 2;
    s = a[m] , l1 = a[m] , r1 = a[m + 1];
    for(int i = m - 1; i >= l; i--){
        s += a[i];
        l1 = max(l1 , s);
    }
    int s1 = a[m + 1];
    for(int i = m + 2; i <= r; i++){
        s1 += a[i];
        r1 = max(r1 , s1);
    }
    return max(max(fen(l , m) , fen(m + 1 , r)) , l1 + r1);
}
bool check(int k){
    for(int i = 1; i <= n; i++){
        que.push(a[i]);
    }
    while(!que.empty()){
        k -= que.top();
        que.pop();
    }
    if(k){
        return true;
    } else {
        return false;
    }
}
int f(int n){
    if(n == 1) return a[1];
    return a[n] + f(n - 1);
}
int main(){
    int n = 2 , ans;
    for(int i = 1; i <= n; i++){
        cin >> a[i];
    }
    ans = fen(1 , n);
    if(check(ans)){
        if(f(n) == ans){
            cout << ans;
        } else {
            cout << "Wrong!";
        }
    } else {
        cout << "Wrong!";
    }
    return 0;
}