FZU2109：Mountain Number(数位DP)

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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

31 101 1001 1000

Sample Output

954384

要求出区间内偶数为大于奇数位的数字的个数

dp[i][j][k] 表示在第i个位置时，前面是j，现在这位是奇数位还是偶数位的数目

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int t;int bit[20],len,l,r;int dp[20][10][2];int dfs(int pos,int pre,int odd,int zero,int doing)//pos-当前位置，pre前一个数字，odd，现在位置的奇偶性，zero是否还是前导0，doing是否美剧导了边界{    if(pos==-1) return 1;    if(dp[pos][pre][odd]!=-1 && !doing)        return dp[pos][pre][odd];    int end = doing?bit[pos]:9;    int ans = 0;    for(int i = 0; i<=end; i++)    {        if(!(i||zero))            ans+=dfs(pos-1,9,0,zero||i,doing&&i==end);        else if(odd && pre<=i)            ans+=dfs(pos-1,i,!odd,zero||i,doing&&i==end);        else if(!odd && pre>=i)            ans+=dfs(pos-1,i,!odd,zero||i,doing&&i==end);    }    if(!doing)        dp[pos][pre][odd] = ans;    return ans;}int cal(int x){    len = 0;    while(x)    {        bit[len++] = x%10;        x/=10;    }    return dfs(len-1,9,0,0,1);//刚刚开始在前面加个9，方便后面判断}int main(){    scanf("%d",&t);    while(t--)    {        memset(dp,-1,sizeof(dp));        scanf("%d%d",&l,&r);        printf("%d\n",cal(r)-cal(l-1));    }    return 0;}

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