BZOJ 1016 Windy 数 | 数位DP

题目:

http://www.lydsy.com/JudgeOnline/problem.php?id=1026

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题解:

f[i][j][1/0]表示枚举到第i位,这位开头是j,当前的数大于(1)或小于(0)目标数的Windy数个数

瞎JB转移即可

#include<cstdio>
#include<algorithm>
#include<cstring>
typedef long long ll;
using namespace std;
ll f[][][],sum[],n,m;
ll dp(int x)
{
    int a[],n=,ret=;
    while (x>) a[++n]=x%,x/=;
    if (n==) a[++n]=;
    memset(f,,sizeof(f));
    for (int i=;i<=;i++)
    if (i<=a[]) f[][i][]=;
    else f[][i][]=;
    for (int i=;i<=n;i++)
    for (int j=;j<=;j++)
        for (int k=;k<=;k++)
        if (abs(j-k)>=)
        {
            if (j<a[i])
            f[i][j][]+=f[i-][k][]+f[i-][k][];
            else if (j==a[i])
            f[i][j][]+=f[i-][k][],f[i][j][]+=f[i-][k][];
            else f[i][j][]+=f[i-][k][]+f[i-][k][];
        }
    for (int i=;i<=;i++)
    if (i<a[n]) ret+=f[n][i][]+f[n][i][];
    else if (i==a[n]) ret+=f[n][i][];
    for (int i=n-;i;i--)
    for (int j=;j<=;j++)
        ret+=f[i][j][]+f[i][j][];
    return ret;
}
int main()
{
    scanf("%lld%lld",&n,&m);
    printf("%lld",dp(m)-dp(n-));
    return ;
}