【Luogu2142】【模板】高精度减法

problem

solution

codes

//高精减法
//高精度的本质，用长度无限的数组或字符串代替int
//原理：模拟减法退位
#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
const int maxn = (int)1e6+10;
int a[maxn],b[maxn],c[maxn], flag;
int main(){
    string s1, s2;
    cin>>s1>>s2;
    if(s1 == s2){ cout<<0<<endl; return 0; }
    if(s1.size()<s2.size() || (s1.size()==s2.size()&&s1<s2)){ swap(s1,s2); flag = 1; }//教训，字符串比较只比较字典序
    a[0] = s1.size(); b[0] = s2.size();
    for(int i = 1; i <= a[0]; i++)a[i] = s1[a[0]-i]-'0';
    for(int i = 1; i <= b[0]; i++)b[i] = s2[b[0]-i]-'0';
    c[0] = max(a[0],b[0])+1;
    for(int i = 1; i <= c[0]; i++){
        c[i] += a[i]-b[i];
        if(c[i] < 0){
            c[i] += 10;
            c[i+1]--;
        }
    }
    while(c[0]>1 && c[c[0]]==0)c[0]--;
    if(flag)cout<<"-";
    for(int i = c[0]; i >= 1; i--)
        cout<<c[i];
    cout<<"
";
    return 0;
}