数位DP

用HDU的一道题目作为实例不要62，这道题是不带前导零的一道实例。

#include <bits/stdc++.h>
using namespace std;

int dp[20][2];
int a[20];

int dfs(int pos, int pre, int state, bool limit)
{
    if (pos == -1) return 1;
    if (!limit && dp[pos][state] != -1) return dp[pos][state];
    int up = limit ? a[pos] : 9;
    int ans = 0;
    
    for (int i = 0; i <= up; ++ i )
    {
        if (pre == 6 && i == 2) continue;
        if (i == 4) continue;
        
        ans += dfs(pos-1, i, i == 6, limit && i == a[pos]);
    }
    if (!limit) dp[pos][state] = ans;
    return ans;
}

int solve(int x)
{
    int pos = 0;
    while (x)
    {
        a[pos++] = x%10;
        x /= 10;
    }
    return dfs(pos-1, -1, 0, true);
}

int main()
{
    int l, r;
    while (scanf("%d %d", &l, &r) != EOF && l+r)
    {
        memset(dp, -1, sizeof(dp));
        printf("%d\n", solve(r)-solve(l-1));
    }
    
    return 0;
}

然后用另一道包含前导零的题，HDU.

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;

int dp[32][66];
int a[66];

int dfs(int pos, int state, bool lead, bool limit)
{
    if (pos == -1) return state >= 32;
    if (!limit && !lead && dp[pos][state] != -1) return dp[pos][state];
    int up = limit ? a[pos] : 1;
    int ans = 0;
    
    for (int i = 0; i <= up; ++ i )
    {
        if (lead && i == 0) ans += dfs(pos-1, state, lead, limit && i == a[pos]);
        else ans += dfs(pos-1, state+(i == 0 ? 1 : -1), lead && i == 0, limit && i == a[pos]);
    }
    if (!limit && !lead) dp[pos][state] = ans;
    return ans;
}

int solve(int x)
{
    int pos = 0;
    while (x)
    {
        a[pos++] = x&1;
        x >>= 1;
    }
    return dfs(pos-1, 32, true, true);
}

int main()
{
    int l, r;
    memset(dp, -1, sizeof(dp));
    while (scanf("%d %d", &l, &r) != EOF)
    {
        // memset(dp, -1, sizeof(dp));
        printf("%d\n", solve(r)-solve(l-1));
    }
    
    return 0;
}

来源：CSDN

作者：Lannnnh.

链接：https://blog.csdn.net/qq_39567491/article/details/104027887

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数位dp