思路

简单的数位DP题.
预处理出表示位,最高位为的Windy数有多少.
然后将范围转换为前缀和(这是有多套路qwq),从高位枚举到低位,加上选小于当前位的数的合法方案数,如果当前位到顶,继续枚举低位.然后别忘了最后答案+1.(一切都是多么经典qwq)
算法复杂度为.

代码

#include<bits/stdc++.h>
using namespace std;
#define i64 long long
#define fp( i, b, e ) for ( int i(b), I(e); i <= I; ++i )
#define fd( i, b, e ) for ( int i(b), I(e); i >= I; --i )
#define go( i, b ) for ( int i(b), v(to[i]); i; v = to[i = nxt[i]] )
template<typename T> inline void cmax( T &x, T y ){ x < y ? x = y : x; }
template<typename T> inline void cmin( T &x, T y ){ y < x ? x = y : x; }

clock_t t_bg, t_ed;
int f[15][15], w[15], n;
int calc( int x ){
    if ( !x ) return 0;
    n = 0; while( x ) w[++n] = x % 10, x /= 10;
    int ans(0), c(88); fd( i, n - 1, 1 ) fp( j, 1, 9 ) ans += f[i][j];
    fd( i, n, 1 ){
        fp( j, i == n, w[i] - 1 ){
            if ( abs(j - c) > 1 ){
                ans += f[i][j];
            }
        } if ( abs(w[i] - c) > 1 ) c = w[i]; else return ans;
    } return ans + 1;
}

signed main(){
    t_bg = clock();
    fp( i, 0, 9 ) f[1][i] = 1;
    fp( i, 2, 10 ) fp( j, 0, 9 ){
        fp( k, 0, j - 2 ) f[i][j] += f[i - 1][k];
        fp( k, j + 2, 9 ) f[i][j] += f[i - 1][k];
    } int x, y; scanf( "%d%d", &x, &y ), printf( "%d\n", calc(y) - calc(x - 1) );
    t_ed = clock();
    fprintf( stderr, "\n========info========\ntime : %.3f\n====================\n", (double)( t_ed - t_bg ) / CLOCKS_PER_SEC );
    return 0;
}