数字计数(数位DP板子题)

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已于 2023-01-05 12:02:46 修改

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文章标签：算法数位DP

于 2023-01-05 11:40:18 首次发布

本文链接：https://blog.youkuaiyun.com/gkajvsbja/article/details/128560531

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该篇文章提供了一个C++代码实现，用于解决洛谷P2602问题，主要涉及动态规划方法来计算在给定范围内，数字的某一位上出现的次数。代码中定义了函数`dfs`进行递归计算，并通过`run`函数处理输入和输出。

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//https://www.luogu.com.cn/problem/P2602
#include <bits/stdc++.h>
#define ll long long
#define all(a) (a).begin(), (a).end()
#define IOS ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
using namespace std;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int N = 1e5 + 10;

int num[20];
ll pow10[20], now[20];
ll f[20];

ll dfs(int pos, int dig, bool lead, bool limit)
{
    if(pos == 0) return 0;
    if(!lead && !limit && f[pos] != -1) return f[pos];
    ll ans = 0;
    int up = limit ? num[pos] : 9;
    for(int i = 0; i <= up; i++)
    {
        if(lead && i == 0) 
            ans += dfs(pos - 1, dig, lead && i == 0, limit && i == up);
        else if(i == dig && limit && i == up)
            ans += now[pos - 1] + 1 + dfs(pos - 1, dig, lead && i == 0, limit && i == up);
        else if(i == dig)
            ans += pow10[pos - 1] + dfs(pos - 1, dig, lead && i == 0, limit && i == up);
        else
            ans += dfs(pos - 1, dig, lead && i == 0, limit && i == up);
    }
    if(!lead && !limit) f[pos] = ans;
    return ans;
}

ll run(ll x, int dig)
{
    int len = 0;
    while(x)
    {
        num[++len] = x % 10;
        x /= 10;
        now[len] = now[len - 1] + pow10[len - 1] * num[len];
    }
    memset(f, -1, sizeof f);
    return dfs(len, dig, true, true);
}

void solve()
{
    ll l, r;
    cin >> l >> r;

    pow10[0] = 1;
    for(int i = 1; i <= 15; i++) pow10[i] = pow10[i - 1] * 10;

    for(int i = 0; i <= 9; i++) cout << run(r, i) - run(l - 1, i) << " \n"[i == 9];
}

signed main()
{
    IOS;
    int t = 1;
    // cin >> t;
    while (t--)
        solve();
    return 0;
}