SPOJ10606 BALNUM - Balanced Numbers(數位DP+狀壓)

Balanced numbers have been used by mathematicians for centuries. A positive integer is considered a balanced number if:

1)      Every even digit appears an odd number of times in its decimal representation

2)      Every odd digit appears an even number of times in its decimal representation

For example, 77, 211, 6222 and 112334445555677 are balanced numbers while 351, 21, and 662 are not.

Given an interval [A, B], your task is to find the amount of balanced numbers in [A, B] where both A and B are included.

Input

The first line contains an integer T representing the number of test cases.

A test case consists of two numbers A and B separated by a single space representing the interval. You may assume that 1 <= A <= B <= 10¹⁹ 

Output

For each test case, you need to write a number in a single line: the amount of balanced numbers in the corresponding interval

Example

Input: 2 1 1000 1 9

Output: 147 4

題意：求l-r之間13579是偶數個，24680是奇數個的數的個數

題解：狀壓壓一下每一位是奇是偶，1表示奇，2表示偶，0表示沒取
dp[pos][sta]表示第pos位以前sta的數有幾個
最基礎的數位DP寫法
記得去一下前導零

代碼以下：

#include<bits/stdc++.h>
using namespace std; int n; long long l,r; long long dp[23][60000][2],a[23],b3[12]; int gg(int x,int pos) { return (x%b3[pos+1])/b3[pos]; } inline int check(int sta) { for(int i=1;i<=9;i+=2) { if(gg(sta,i)==1) return 0; } for(int i=0;i<=8;i+=2) { if(gg(sta,i)==2) return 0; } return 1; } long long dfs(int pos,int sta,int lim,int lim2) { if(pos<=0) return check(sta); if(!lim&&dp[pos][sta][lim2]!=-1) return dp[pos][sta][lim2]; int up=lim?a[pos]:9; long long res=0; int nextsta; for(int i=0;i<=up;i++) { if(!lim2&&i==0) { res+=dfs(pos-1,sta,lim&&i==a[pos],lim2); } else { if(gg(sta,i)!=2) nextsta=sta+b3[i]; else nextsta=sta-b3[i]; res+=dfs(pos-1,nextsta,lim&&i==a[pos],lim2|1); } } if(!lim) dp[pos][sta][lim2]=res; return res; } long long get(long long x) { memset(dp,-1,sizeof(dp)); int cnt=0; while(x) { a[++cnt]=x%10; x/=10; } return dfs(cnt,0,1,0); } int main() { b3[0]=1; for(int i=1;i<=11;i++) b3[i]=b3[i-1]*3; scanf("%d",&n); while(n--) { scanf("%lld%lld",&l,&r); printf("%lld\n",get(r)-get(l-1)); } }