noip模拟赛 终末

分析：举个例子就能发现：偶数位上的数都必须是0，奇数位上的数可以取0~k-1,这就是一个标准的数位dp了.

      这编译器......数组越界了竟然不报错.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long ll;
ll n, k, shu[2000000], kk, f[200000];

ll dfs(ll len, bool limit)
{
    if (len == 0)
        return 1;
    if (!limit && f[len])
        return f[len];
    ll cnt = 0, maxx = (limit ? shu[len] : k - 1);
    
    if (len % 2 != 0)
    {
        for (int i = 0; i <= maxx; i++)
            cnt += dfs(len - 1, limit && i == maxx);
        return limit ? cnt : f[len] = cnt;
    }
    else
    {
        cnt += dfs(len - 1, limit && maxx == 0);
            return limit ? cnt : f[len] = cnt;
    }
    
}

ll solve(ll x)
{
    while (x)
    {
        shu[++kk] = x % k;
        x /= k;
    }
    return dfs(kk, true);
}

int main()
{
    scanf("%lld%lld", &n, &k);
    printf("%lld\n", solve(n));

    return 0;
}

 

转载于:https://www.cnblogs.com/zbtrs/p/7767180.html