HDU 5327 Olympiad（转）（数组前缀和）

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

#include <stdio.h>
#include <string.h>
#define N 100010
int a[10], f[N];

int check(int m) {
	memset(a, 0, sizeof(a));
	while (m) {
		if (a[m % 10])
			return 0;
		else
			a[m % 10] = 1;
		m /= 10;
	}
	return 1;
}

void pre() {
	//f[1] = 1;
	for (int i = 1; i < N; i++)
		f[i] = f[i - 1] + check(i);
}

int main() {
	int n, a, b;
	pre();
	scanf("%d", &n);
	while (n--) {
		scanf("%d %d", &a, &b);
		printf("%d\n", f[b] - f[a - 1]);
	}
	return 0;
}

转自：http://blog.youkuaiyun.com/jtjy568805874/article/details/47154613

#include<sstream>
#include<string>
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include <iterator>
#include<vector>
#include<map>
#include <stack>
#include<queue>
#include<set>
#include <list>
#include<functional>
#include<numeric>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
inline int lowbit(int x){ return x&(-x);}
typedef  long long int LL;
const int INF = 0x3f3f3f3f ;
const double eps = 1e-6;
const long double PI = acos(0.0) * 2.0;
const int N = 100009;
int sum[N];
bool check(int x);

int main()
{
    sum[0] = 0;
    for(int i = 1 ; i <= N ;i++)
        sum[i] = check(i) ? 1 : 0; //标记是不是美丽数
    for(int  i = 2 ; i <= N ;i++)
        sum[i] += sum[i-1];    //计算前缀和

    int Case;
    scanf("%d",&Case);
    while(Case--)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        printf("%d\n",sum[b]-sum[a-1]);
    }
    return 0;
}

bool check(int x)
{
    set<int> s;
    while(x)
    {
        int  y = x%10;
        if(!s.count(y))
            s.insert(y);
        else
            return false;
        x/=10;
    }
    return true;
}

转自：http://blog.youkuaiyun.com/tt2767