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  • AcWing 1081. 度的数量

    • 代码:
    #include <algorithm>
    #include <cmath>
    #include <cstdio>
    #include <cstring>
    #include <iostream>
    #include <map>
    #include <queue>
    #include <vector>
    
    using namespace std;
    typedef long long ll;
    const ll N = 33;
    const ll mod = 1000000007;
    ll ans = 0;
    ll K, B;
    ll dp[N][N];
    ll C[N][N];
    void init() {
        for (int i = 0; i < N; i++) {
            for (int j = 0; j <= i; j++) {
                if (j == 0)
                    C[i][j] = 1;
                else
                    C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
            }
        }
    }
    ll DP(ll n) {
        if (n == 0) return 0;
        vector<int> nums;
        ll now = n;
        while (now) {
            nums.push_back(now % B);
            now /= B;
        }
        ll res = 0, last = 0;
        for (int i = nums.size() - 1; i >= 0; i--) {
            if (nums[i] > 0) {
                res += C[i][K - last];
                if (nums[i] > 1) {
                    if (K - last - 1 >= 0) res += C[i][K - last - 1];
                    break;
                } else {
                    last++;
                    if (last > K) break;
                }
            }
            if (!i && last == K) res++;
        }
        return res;
    }
    void solve() {
        ll a, b;
        cin >> a >> b >> K >> B;
        cout << DP(b) - DP(a - 1) << endl;
    }
    signed main() {
        ios::sync_with_stdio(0);
        ll t = 1;
        init();
        while (t--) solve();
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/Xiao-yan/p/14595577.html
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