A .Gaby And Addition （Gym

题目链接：http://codeforces.com/gym/101466/problem/A

题目：

题意：

　　给你n个数，重定义两个数之间的加法不进位，求这些数中两个数相加的最大值和最小值。

思路：

　　字典树。我们首先将前i-1为放入字典树中，然后在查询第i位时，我们去字典树中查询，对每一位进行寻找，找到满足题意的当前位的最大值和最小值，然后继续更新下一位，最后维护总的最大值和最小值即可。

代码实现如下：

#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<ll, int> pli;
typedef pair<int, ll> pil;;
typedef pair<int, int> pii;
typedef unsigned long long ull;

#define lson i<<1
#define rson i<<1|1
#define lowbit(x) x&(-x)
#define bug printf("*********
");
#define debug(x) cout<<"["<<x<<"]" <<endl;
#define FIN freopen("D://code//in.txt", "r", stdin);
#define IO ios::sync_with_stdio(false),cin.tie(0);

const double eps = 1e-8;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;

int n;
int le,root;
int arr[20];
ll num, pw[20];

struct node{
    int nxt[10];
    void init(){
        for(int i = 0; i < 10; i++) nxt[i] = -1;
    }
}T[13*maxn];

void insert(ll x){
    int now = root;
    for(int i = 0; i <= 18; i++) {
        arr[i] = x % 10;
        x /= 10;
    }
    for(int i = 18;i >= 0; i--){
        int num = arr[i];
        if(T[now].nxt[num] == -1){
            T[le].init();
            T[now].nxt[num] = le++;
        }
        now = T[now].nxt[num];
    }
}

ll search1(ll x){
    int now = root, mx, idx;
    ll res = 0;
    for(int i = 0; i <= 18; i++) {
        arr[i] = x % 10;
        x /= 10;
    }
    for(int i = 18; i >= 0; i--) {
        mx = -1, idx = -1;
        for(int j = 0; j < 10; j++) {
            if(T[now].nxt[j] != -1 && (j + arr[i]) % 10 > mx) {
                mx = (j + arr[i]) % 10;
                idx = j;
            }
        }
        now = T[now].nxt[idx];
        res = res + mx * pw[i];
    }
    return res;
}

ll search2(ll x){
    int now = root, mx, idx;
    ll res = 0;
    for(int i = 0; i <= 18; i++) {
        arr[i] = x % 10;
        x /= 10;
    }
    for(int i = 18; i >= 0; i--) {
        mx = 11, idx = -1;
        for(int j = 0; j < 10; j++) {
            if(T[now].nxt[j] != -1 && (j + arr[i]) % 10 < mx) {
                mx = (j + arr[i]) % 10;
                idx = j;
            }
        }
        now = T[now].nxt[idx];
        res = res + mx * pw[i];
    }
    return res;
}

int main() {
    le = 1;
    pw[0] = 1;
    for(int i = 1; i <= 18; i++) pw[i] = pw[i-1] * 10;
    T[0].init();
    scanf("%d", &n);
    ll ans1 = INF, ans2 = -1;
    for(int i = 1; i <= n; i++) {
        scanf("%lld", &num);
        if(i > 1) {
            ans1 = min(search2(num), ans1);
            ans2 = max(search1(num), ans2);
        }
        insert(num);
    }
    printf("%lld %lld
", ans1, ans2);
    return 0;
}