# BZOJ1026 [SCOI2009]windy数(数位DP)

/**************************************************************
Problem: 1026
Language: C++
Result: Accepted
Time:60 ms
Memory:1288 kb
****************************************************************/

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>

#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x))

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-6;
const int mod = 100003;
const int maxn = 2e5+100;
const int maxm = 2e5+100;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const db pi = acos(-1.0);

int f[20][20];//首位为j的i位数的所有数中windy数的个数
int sum[20];//不含前导零的i位windy数
int st[20];//数的每一位
int top;

int solve(int x){
int ans = 0;
int top = 0;
int p = 1;
while(x){
st[++top] = x%10;
ans += sum[p-1];
p++;
x /= 10;
}

for(int i = 1; i < st[top]; i++){
ans += f[top][i];
}
for(int i = top-1; i > 0; i--){
for(int j = 0; j < st[i]; j++){
if(abs(j-st[i+1])<2)continue;
ans += f[i][j];
}
if(abs(st[i]-st[i+1])<2)break;
}
return ans;
}

int main() {
int A, B;
scanf("%d %d", &A, &B);
mem(f, 0);
mem(sum, 0);
for(int i = 0; i < 10; i++) f[1][i] = 1;
for(int i = 2; i <= 10; i++){
for(int j = 0; j < 10; j++){
for(int k = 0; k < 10; k++){
if(abs(j-k)<2)continue;
f[i][j] += f[i-1][k];
}
}
}
for(int i = 1; i <= 10; i++){
for(int j = 1; j <= 10; j++){
sum[i] += f[i][j];
}
}
printf("%d", solve(B+1)-solve(A));
return 0;
}

posted @ 2018-10-02 22:43  wrjlinkkkkkk  阅读(137)  评论(0编辑  收藏