FZU2109 Mountain Number (数位DP)

Problem 2109 Mountain Number
Accept: 213 Submit: 538
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

#include "cstring"#include "iostream"#include "string.h"#include "cstdio"using namespace std;int bit[20],len;int dp[20][10][2];// pos表示当前位置,pre表示之前一个数字,odd表示当前是奇数位(0)还是偶数位(1)，zero表示是否是前导0 有前导0(0) 没有前导0(1)，over表示是否越界   over 为1是前一位没有达到上界int dfs(int pos,int pre,int odd,int zero,int over){    if(pos<0)        return 1;    if(dp[pos][pre][odd]!=-1&&over)    {        return dp[pos][pre][odd];    }    int last=over?bit[pos]:9;    int ans=0;    for(int i=0;i<=last;i++)    {        //这个位置为0的同时这个位置有前导0        if( !( i||zero) )            ans += dfs(pos-1,i,!odd,zero||i,over&&i==last);        //如果是偶数位的同时当前值大于pre        else if(odd && pre<=i)            ans += dfs(pos-1,i,!odd,zero||i,over&&i==last);        //如果是奇数位的同时当前项小于pre        else if(!odd && pre>=i)            ans += dfs(pos-1,i,!odd,zero||i,over&&i==last);    }    if(!over)        dp[pos][pre][odd]=ans;    return ans;}int solve(int x){    len=0;    while(x)    {        bit[len++]=x%10;        x/=10;    }    return dfs(len-1,9,0,0,1);}int main(){    int cas;    scanf("%d",&cas);    while(cas--)    {        int l,r;        memset(dp,-1,sizeof(dp));        scanf("%d%d",&l,&r);        printf("%d\n",solve(r)-solve(l-1));    }    return 0;}