2015多校联合第四场hdu5327Olympiad 递推

Problem Description

You are one of the competitors of the Olympiad in numbers. The problem of this year relates to beatiful numbers. One integer is called beautiful if and only if all of its digitals are different (i.e. 12345 is beautiful, 11 is not beautiful and 100 is not beautiful).
Every time you are asked to count how many beautiful numbers there are in the interval [a,b] (a≤b).
Please be fast to get the gold medal!



Input

The first line of the input is a single integer T (T≤1000),
indicating the number of testcases.

For each test case, there are two numbers a and b,
as described in the statement. It is guaranteed that 1≤a≤b≤100000.



Output

For each testcase, print one line indicating the answer.



Sample Input

2
1 10
1 1000

Sample Output

10
738

最开始的思路是特别暴力的逐个试，超时，搜题解，有这么做的，更奇葩的是人家用cin cout AC了，我是scanf 把他（按比例来说更大些）的代码换成scanf printf超时了……什么鬼，（我为了拉低杭电AC率做出了卓越贡献

）

自己找别的思路，可能有公式？推导半天没有头绪，始终觉得预处理可能会好些 果不其然 而且是0ms的代码 太优美了~

#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
bool num[10],flag;
int t,a,b,m,n,f[100005];
bool check(int x)
{
    memset(num,0,sizeof(num));
    for(int i=0;x;x/=10)
    {
        if(num[x%10]) return 0;
        else num[x%10]=1;
    }
    return 1;
}
void init()
{
    for(int i=1;i<=100000;i++)
        f[i]=f[i-1]+check(i);//这里的递推好赞
}
int main()
{
    init();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        printf("%d\n",f[m]-f[n-1]);
    }
    return 0;
}