P2657 [SCOI2009]windy数

P2657 [SCOI2009]windy数

题目描述

windy定义了一种windy数。不含前导零且相邻两个数字之差至少为2的正整数被称为windy数。 windy想知道，

在A和B之间，包括A和B，总共有多少个windy数？

Solution

有先导 \(0\) 的数位\(dp\)

把此位前有无前导 \(0\) 作为搜索的一个状态即可

注意有前导 \(0\) 时不能直接返回， 因为有前导 \(0\) 就代表着无法到达 \(10^{len} - 1\)

Code

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#include<climits>
#define LL long long
#define REP(i, x, y) for(LL i = (x);i <= (y);i++)
using namespace std;
LL RD(){
    LL out = 0,flag = 1;char c = getchar();
    while(c < '0' || c >'9'){if(c == '-')flag = -1;c = getchar();}
    while(c >= '0' && c <= '9'){out = out * 10 + c - '0';c = getchar();}
    return flag * out;
    }
const LL maxn = 19;
LL num[maxn];
LL dp[maxn][maxn];
LL DP(LL Index, LL state, LL zero, bool limit){
	if(Index == 0)return 1;
	if(!zero && !limit && dp[Index][state] != -1)return dp[Index][state];
	LL ans = 0, up = limit ? num[Index] : 9;
	REP(i, 0, up){
		if(zero)ans += DP(Index - 1, i, i == 0, limit && (i == num[Index]));
		else{
			if(abs(i - state) < 2)continue;
			ans += DP(Index - 1, i, 0, limit && (i == num[Index]));
			}
		}
	if(!zero && !limit)dp[Index][state] = ans;
	return ans;
	}
LL solve(LL x){
	LL len = 0;
	while(x){
		num[++len] = x % 10;
		x /= 10;
		}
	return DP(len, 0, 1, 1);
	}
int main(){
	memset(dp, -1, sizeof(dp));
	LL l = RD(), r = RD();
	printf("%lld\n", solve(r) - solve(l - 1));
	return 0;
	}