高精度 AcWing 792. 高精度减法

川长思鸟来 2023-09-28 11:11 143阅读 0赞

## 高精度 AcWing 792. 高精度减法 ##

### 原题链接 ###

[AcWing 792. 高精度减法][AcWing 792.]

### 算法标签 ###

高精度

### 代码 ###

#include<bits/stdc++.h>
    #define int long long
    #define rep(i, a, b) for(int i=a;i<b;++i)
    #define Rep(i, a, b) for(int i=a;i>b;--i)
    using namespace std;
    const int N = 100005;
    int a[N];
    inline int read(){
       int s=0,w=1;
       char ch=getchar();
       while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
       while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
       return s*w;
    }
    void put(int x) {
        if(x<0) putchar('-'),x=-x;
        if(x>=10) put(x/10);
        putchar(x%10^48);
    }
    bool cmp(vector<int> &A, vector<int> &B){
        if (A.size() != B.size()) return A.size() > B.size();
        for (int i = A.size() - 1; i >= 0; i -- )
            if (A[i] != B[i])
                return A[i] > B[i];
        return true;
    }
    vector<int> sub(vector<int> &a, vector<int> &b){
        vector<int> c;
        int t=0;
        rep(i, 0, a.size()){
            t=a[i]-t;
            if(i<b.size()){
                t-=b[i];
            }
            c.push_back((t+10)%10);
            if(t<0){
                t=1;
            }else{
                t=0;
            }
        }
        while(c.size()>1&&c.back()==0){
            c.pop_back();
        }
        return c;
    }
    signed main(){
    	ios::sync_with_stdio(false);
    	cin.tie(0);
    	cout.tie(0);
    	string a,b;
    	cin>>a>>b;
    	vector<int> va, vb;
    	Rep(i, a.size()-1, -1){
    	    va.push_back(a[i]-'0');
    	}
    	Rep(i, b.size()-1, -1){
    	    vb.push_back(b[i]-'0');
    	}
    	vector<int> c;
    	if(cmp(va, vb)){
    	    c=sub(va, vb);
    	}
    	else if(!cmp(va, vb)){
    	    c=sub(vb, va);
    	    printf("-");
    	}
    	Rep(i, c.size()-1, -1){
    	    printf("%lld", c[i]);
    	}
    	return 0;
    }

### 模板 ###

bool cmp(vector<int> &A, vector<int> &B){
        if (A.size() != B.size()) return A.size() > B.size();
        for (int i = A.size() - 1; i >= 0; i -- )
            if (A[i] != B[i])
                return A[i] > B[i];
        return true;
    }
    // C = A - B, 满足A >= B, A >= 0, B >= 0
    vector<int> sub(vector<int> &A, vector<int> &B){
        vector<int> C;
        for (int i = 0, t = 0; i < A.size(); i ++ ){
            t = A[i] - t;
            if (i < B.size()) t -= B[i];
            C.push_back((t + 10) % 10);
            if (t < 0) t = 1;
            else t = 0;
        }
        while (C.size() > 1 && C.back() == 0) C.pop_back();
        return C;
    }

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![在这里插入图片描述][9cb4cb80a18345158f8a69e2f196e5b9.png]

[AcWing 792.]: https://www.acwing.com/problem/content/794/
[9cb4cb80a18345158f8a69e2f196e5b9.png]: https://img-blog.csdnimg.cn/9cb4cb80a18345158f8a69e2f196e5b9.png