P2657 [SCOI2009] windy 数

原题链接

• 代码：

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <map>
#include <queue>
#include <vector>

using namespace std;
typedef long long ll;
const ll N = 15;
const ll mod = 1000000007;
ll ans = 0;
ll K, B;
ll f[N][N];
void init() {
    for (int i = 0; i <= 9; i++) {
        f[1][i] = 1;
    }
    for (int i = 2; i < N; i ++) {
        for (int j = 0; j <= 9; j ++) {
            for (int k = 0; k <= 9; k ++) {
                if (abs(k-j)>= 2)f[i][j] += f[i-1][k];
            }
        }
    }
}
ll DP(ll n) {
    /******************************模板*********/
    if (n == 0) return 0;//这个也是看情况返回0还是个1
    vector<int> nums;
    ll now = n;
    while (now) {
        nums.push_back(now % 10);
        now /= 10;
    }
    ll res = 0, last = -4;//这个last是看情况搞
    /******************************模板*********/
    for (int i = nums.size() - 1; i >= 0; i--) {
        ll x = nums[i];
        for (int j = (i == nums.size()-1)/*是因为非最高位的话此时可以为0，如果是最高位，因为没前导0，所以为1*/; j < x; j ++) {
            if (abs(j - last) >= 2)//正常判断
            res += f[i+1][j];
        }
        if (abs(last-x) < 2)break;//原数左面是否合法
        last = x;
        if (!i)res++;//套路
    }
    for (int i = 1; i < nums.size(); i++) {
        for (int j = 1; j <= 9; j ++) {//处理没有前导0的情况，
            res += f[i][j];//
        }
    }
    return res;
}
void solve() {
    init();
    ll l, r;
    while (cin >> l >> r)cout << DP(r) - DP(l-1)<< endl;
}
signed main() {
    ios::sync_with_stdio(0);
    ll t = 1;
    while (t--) solve();
    return 0;
}