http://acm.hdu.edu.cn/showproblem.php?pid=5327
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 $[a,b]\ (a \le b)$. Please be fast to get the gold medal!
The first line of the input is a single integer $T\ (T \leq 1000)$, 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 $1 \leq a \leq b \leq 100000$.
For each testcase, print one line indicating the answer.
2
1 10
1 1000
10
738
暴力枚举，前缀和。。
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<map>
using std::map;
using std::min;
using std::find;
using std::pair;
using std::vector;
using std::multimap;
#define pb(e) push_back(e)
#define sz(c) (int)(c).size()
#define mp(a, b) make_pair(a, b)
#define all(c) (c).begin(), (c).end()
#define iter(c) __typeof((c).begin())
#define cls(arr, val) memset(arr, val, sizeof(arr))
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for(int i = 0; i < (int)n; i++)
#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)
const int N = 100001;
const int INF = 0x3f3f3f3f;
int vis[10], sum[N + 10];
void init() {
int v, j;
sum[0] = 0;
for(int i = 1; i <= N; i++) {
v = i;
bool f = true;
cls(vis, 0);
do vis[v % 10]++; while(v /= 10);
for(j = 0; j < 10; j++) {
if(vis[j] > 1) { f = false; break;}
}
sum[i] = sum[i - 1] + f;
}
}
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w+", stdout);
#endif
init();
int t, x, y;
scanf("%d", &t);
while(t--) {
scanf("%d %d", &x, &y);
printf("%d\n", sum[y] - sum[x - 1]);
}
return 0;
}
原文：http://www.cnblogs.com/GadyPu/p/4796277.html