牛客小白月赛7 明七暗七

最新推荐文章于 2024-09-21 10:59:26 发布

weixin_33860147

最新推荐文章于 2024-09-21 10:59:26 发布

阅读量206

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/widsom/p/9703185.html

明七暗七

思路：

二分+数位dp

代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL unsigned long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define piii pair<pii, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

LL dp[20][2][7];
int a[20];
LL dfs(int pos, bool limit, bool appear, int cnt) {
    if(pos == -1) {
        if(appear || cnt == 0) return 1;
        else return 0;
    }
    if(!limit && ~dp[pos][appear][cnt]) return dp[pos][appear][cnt];
    int up = 9;
    if(limit) up = a[pos];
    LL ans = 0;
    for (int i = 0; i <= up; i++) {
        ans += dfs(pos-1, limit&&i==up, appear||i==7, (cnt*10 + i) % 7);
    }
    if(!limit) dp[pos][appear][cnt] = ans;
    return ans;
}

LL solve(LL n) {
    if(n == 0) return 1;
    int cnt = 0;
    mem(dp, -1); 
    while(n) {
        a[cnt++] = n % 10;
        n /= 10;
    }
    return dfs(cnt-1, 1, 0, 0);
}
int main() {
    LL n, m;
    scanf("%lld %lld", &m, &n);
    LL l = m+1, r = 1e18, mid = l+r >> 1;
    LL sub = solve(m);
    while(l < r) {
        if(solve(mid) - sub >= n) r = mid;
        else l = mid + 1;
        mid = l+r >> 1;
    } 
    printf("%lld\n", mid);
    return 0;
}