2019-08-21 纪中NOIP模拟赛A组

T1 [JZOJ6315] 数字

题目描述

数据范围

　　

分析

　　我叕正解写挂了，写模拟一定要仔细考虑细节啊

　　正解是这样说的

　　我是在 $n$ 串中按位数从小到大枚举数字匹配，思路和正解相似，时间复杂度相同，做法稍有不同

#include <iostream>
#include <cstdio>
#include <cstdlib> 
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
#define ll long long

ll inf = 1e18, p[20] = {1};
int t, n, a[20];
char s[20];

ll solve() {
    for (int i = 1; i <= n; i++) {
        ll ans = inf;
        for (int j = 1; j <= i; j++) {
            if (!a[j]) continue;
            ll now = 0; int len = i, k = j + i, f = 0;
            for (int l = 0; l < i; l++) {
                if (j + l <= n) now = now * 10 + a[j + l];
                else now = now * 10 + a[j + l - i];
            }
            if (j + i - 1 > n) {
                ll last = now % p[j + i - 1 - n];
                now = now - last + (last + 1) % p[j + i - 1 - n];
            }
            for (int l = 1; j - l; l++)
                if (a[j - l] != (now - 1) / p[l - 1] % 10) {f = 1; break;}
            if (f) continue;
            if (++now / p[len]) len++;
            while (k + len - 1 <= n) {
                for (int l = 0; l < len; l++)
                    if (a[k + l] != now / p[len - l - 1] % 10) {f = 1; break;}
                if (f) break; k += len;
                if (++now / p[len]) len++;
            }
            if (f) continue;
            if (k > n) now--;
            else {
                for (int l = 0; k + l <= n; l++)
                    if (a[k + l] != now / p[len - l - 1] % 10) {f = 1; break;}
                if (f) continue;
            }
            ans = min(ans, now);
        }
        if (ans < inf) return ans;
    }
}

int main() {
    scanf("%d", &t);
    for (int i = 1; i <= 18; i++) p[i] = p[i - 1] * 10;
    while (t--) {
        scanf("%s", s + 1); n = strlen(s + 1);
        for (int i = 1; i <= n; i++) a[i] = s[i] - '0';
        printf("%lld\n", solve());
    } 
    
    return 0;
}

View Code

 T2 [JZOJ6313] Maja

题目描述