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 简单的状压一下。两个状态。
第一个状态i位上表示第i个数出现的次数是奇数次1，还是偶数次0.
第二个状态i位上表示的是i这个数字有没有出现过，出现1，没出现0.
然后注意一下前导0就OK了。
//#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long LL;
LL l,r;
LL dp[20][1<<10][1<<10];
int digit[20];
int check(int state1,int state2)
{
for(int i = 0 ; i <= 9 ; ++i)
{
int d = 1<<i;
if(state2 & d)
{
if((i%2) == ((state1&d)>0))return 0;
}
}
return 1;
}
LL dfs(int len,int state1,int state2,bool up,bool zero)
{
if(!len)
{
//printf("%d %d %d\n",state1,state2,check(state1,state2));
return check(state1,state2);
}
if(!up && dp[len][state1][state2] != -1)return dp[len][state1][state2];
int n = up?digit[len] : 9;
LL res = 0;
for(int i = 0 ; i <= n ; ++i)
{
if(zero)
{
if(!i)res += dfs(len-1,0,0,up && i == n,1);
else res += dfs(len-1,state1^(1<<i),state2|(1<<i),up && i == n,0);
}
else res += dfs(len-1,state1^(1<<i),state2|(1<<i),up && i == n,0);
}
if(!up)dp[len][state1][state2] = res;
return res;
}
LL cal(LL x)
{
int len = 0;
while(x)
{
digit[++len] = x % 10;
x /= 10;
}
return dfs(len,0,0,1,1);
}
int main()
{
int t;
//printf("%d\n",check(16,16));
scanf("%d",&t);
memset(dp,-1,sizeof dp);
for(int tt = 1 ; tt <= t ; ++tt)
{
scanf("%lld%lld",&l,&r);
printf("%lld\n",cal(r) - cal(l-1));
}
return 0;
}