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USETC 250 windy数 数位DP

注意处理数字只有一位的情况(其实不用怎么处理)= =

简单数位DP

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <climits>#include <string>#include <iostream>#include <map>#include <cstdlib>#include <list>#include <set>#include <queue>#include <stack>using namespace std;typedef long long LL;const int maxn = 20;int lim[maxn],len;int num[maxn];int f[15][15][15][2];void getlim(int num) {    memset(lim,0,sizeof(lim));    len = 0;    while(num) {        lim[len++] = num % 10;        num /= 10;    }}int myabs(int x) {    return x < 0 ? -x : x;}int dfs(int now,int prev,int maxd,int first,int bound) {    int &note = f[now][prev][maxd][first];    if(now == 0) {        if(first && maxd >= 2) {            return 1;        }        return 0;    }    if(!bound && note != -1) return note;    int m = bound ? lim[now - 1] : 9,ret = 0;    for(int i = 0;i <= m;i++) {        num[now - 1] = i;        if(first) {            int nmaxd = min(maxd,myabs(i - prev));            ret += dfs(now - 1,i,nmaxd,1,bound && i == m);        }        else {            ret += dfs(now - 1,i,maxd,i || first,bound && i == m);        }    }    if(!bound) note = ret;    return ret;}int work(int num) {    memset(f,-1,sizeof(f));    if(num < 10) return num;    getlim(num);    return dfs(len,0,11,0,1);}int main() {    int n,m;    while(scanf("%d%d",&n,&m) == 2) {        printf("%d\n",work(m) - work(n - 1));    }    return 0;}