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  • 牛客小白月赛7 明七暗七

    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
    ", mid);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/widsom/p/9703185.html
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