FZU2109 Moutain Number 数位DP

One integer number x is called “Mountain Number” if:
(1) x>0 and x is an integer;
(2) Assumex=a[0]a[1]...a[len−2]a[len−1]x=a[0]a[1]...a[len-2]a[len-1]x=a[0]a[1]...a[len−2]a[len−1](0≤a[i]a[i]a[i]≤9,a[0]a[0]a[0] is positive). Any a[2i+1]a[2i+1]a[2i+1] is larger or equal to a[2i]a[2i]a[2i] and a[2i+2]a[2i+2]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) ?

题意：满足条件的数字是（从左往右开始数为第0位）奇数位不能小于相邻的偶数位。
这样dp就开三维记录前一位数，这一位是奇数位还是偶数位，还有第几位。
在dfs过程中，如果是偶数位就要比前一位小，如果是奇数位就要比前一位大。
还有要注意前导零，因为它要从第一个不是零的数位算起，这一位才是真正的第0位。
代码中QAQ如果是 0代表是偶数位，1代表是奇数位。

#include <cstdio>
#include <cstring>
using namespace std;

int dp[20][20][2], a[25];
int dfs(int pos, bool limit, int pre, int QAQ, bool QianDao0){
if(pos < 0) return 1;
if(!limit && dp[pos][pre][QAQ] != -1) return dp[pos][pre][QAQ];
int up = limit ? a[pos] : 9, ans = 0;
for(int i = 0; i <= up; i++){
if(QianDao0 && !i) ans += dfs(pos - 1, false, 10, 0, true);
else if(QAQ && i >= pre) ans += dfs(pos - 1, limit && i == up, i, !QAQ, false);
else if(!QAQ && i <= pre) ans += dfs(pos - 1, limit && i == up, i, !QAQ, false);
}
return (!limit) ? dp[pos][pre][QAQ] = ans : ans;
}
int solve(int x){
int pos = 0;
while(x){
a[pos++] = x % 10;
x /= 10;
}
return dfs(pos - 1, true, 10, 0, true);
}
int main(){
int t;
scanf("%d", &t);
memset(dp, -1, sizeof(dp));
while(t--){
int l, r;
scanf("%d%d", &l, &r);
printf("%d\n", solve(r) - solve(l - 1));
}
return 0;
}