• Fzu2109 Mountain Number 数位dp


    Accept: 189    Submit: 461
    Time Limit: 1000 mSec    Memory Limit : 32768 KB

     Problem Description

    One integer number x is called "Mountain Number" if:

    (1) x>0 and x is an integer;

    (2) Assume x=a[0]a[1]...a[len-2]a[len-1](0≤a[i]≤9, a[0] is positive). Any a[2i+1] is larger or equal to a[2i] and a[2i+2](if exists).

    For example, 111, 132, 893, 7 are "Mountain Number" while 123, 10, 76889 are not "Mountain Number".

    Now you are given L and R, how many "Mountain Number" can be found between L and R (inclusive) ?

     Input

    The first line of the input contains an integer T (T≤100), indicating the number of test cases.

    Then T cases, for any case, only two integers L and R (1≤L≤R≤1,000,000,000).

     Output

    For each test case, output the number of "Mountain Number" between L and R in a single line.

     Sample Input

    3 1 10 1 100 1 1000

     Sample Output

    9 54 384

     Source

    “高教社杯”第三届福建省大学生程序设计竞赛
     
    某次群赛的一个很水的数位dp,但脑抽了,不知道想到哪里去了
    #include <cstdio>
    #include <cstring>
    #include <iostream>
    using namespace std;
    typedef long long ll;
    int bits[20];
    ll dp[20][15][2][2];
    ll dfs(int pos, int pre, int iso, int doing) {
        if(pos == -1) return 1;
        int End = doing ? bits[pos] : 9;
        ll& ans = dp[pos][pre][iso][doing];
        if(ans != -1) return ans;
        ans = 0;
        for(int i = 0; i <= End; ++i) {
            if(iso && i <= pre) ans += dfs(pos - 1, i, 0, doing && i == End);
            else if(!iso && i >= pre) ans += dfs(pos - 1, i, 1, doing && i == End);
        }
        return ans;
    }
    ll calc(ll x) {
        int ls = 0;
        memset(dp, -1, sizeof dp);
        while(x) {
            bits[ls++] = x % 10;
            x /= 10;
        }
        return dfs(ls - 1, 9, 1, 1);
    }
    int main() {
        ll n, m;
        int _;
        cin >> _;
        while(_ --) {
            cin >> n >> m;
            cout << calc(m) - calc(n - 1) << endl;
        }
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/orchidzjl/p/5155811.html
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