I. The Crime-solvingPlan of Groundhog (思维 / 高精度乘法) 2020牛客暑期多校训练营(第九场)

传送门

思路：

• 题意： 给出n个数，将其拼接成两个没有前导零的正整数，使得他们的乘积最小。
• 官方题解：
  
• 我想的思路个题解一样，但是我在补题的时候出现了一个小问题(代码中已注释)，导致我一直改其他地方一直wa，呜呜呜~

代码实现：

#include<bits/stdc++.h>
//#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair<int, int>
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
const int  inf = 0x7fffffff;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll   mod = 1e9 + 7;
const int  N = 2e5 + 5;

int t, n, a[N], b;
vector<int> A, C;

int cmp(int a, int b){
    return a > b;
}

vector<int> mul(vector<int> &A, int b)
{
    vector<int> C;
    int t = 0;
    for(int i = 0; i < A.size(); i ++){
        t += A[i] * b;
        C.push_back(t % 10);
        t /= 10;
    } //高精度乘法板子的小瑕疵，这里的t不能放在循环的判断语句里写成(i < A.size() || t)
    while(t) C.push_back(t % 10), t /= 10;
    while(!C.back() && C.size()>1) C.pop_back();
    return C;
}

signed main()
{
    IOS;

    cin >> t;
    while(t --){
        cin >> n;
        int minn = inf, maxx = 0, cnt0 = 0;
        for(int i = 0; i < n; i ++){
            cin >> a[i];
            if(!a[i]) cnt0 ++;
            else minn = min(minn, a[i]);
            maxx = max(maxx, a[i]);
        }
        b = minn; //取出最小的正整数
        sort(a, a+n, cmp);
        int pos = -1; //记录最小的首位数字在a[]中的位置
        for(int i = n-1; ~i; i --) if(a[i] == minn) {pos = i-1;break;}
//        cout << "pos: " << pos << endl;

        if(pos == -1){
            cout << 0 << endl;
            continue;
        }

        bool flag = 0; A.clear();

        for(int i = 0; i < pos; i ++) A.push_back(a[i]);
        while(cnt0 --) A.push_back(0); A.push_back(a[pos]);

//        cout << "A: ";
//        for(int i = 0; i < A.size(); i ++) cout << A[i];
//        cout << endl;
//        cout << "b: " << b << endl;

        if(b > 1) C = mul(A, b);
        else C = A;
        for(int i = C.size()-1; ~i; i --) cout << C[i];
        cout << endl;
    }

    return 0;
}