uva12527 Different Digits

uva12527

uDebug12527

簡單題，一個數字（number）的各個位（digit）上的數字如果有重複，則視爲有重複數位的數。給定兩個數字0<=N<=M<= 5000，求N和M之間沒有重複數位的數有多少個。

由於本身數字也不多，直接開個數組d，把每個數i是否是重複數位的結果存儲到d[i]，然後再另一個數組s，直接累加結果，答案也就很快出來了。

python版本代碼

s = [0 for i in range(5001)]
d = [0 for i in range(5001)]

def IsRepeat(num):
    a = [0 for i in range(10)]
    while num > 0:
        x = num % 10
        if a[x] == 0:
            a[x] = 1
        else:
            return False
        num //= 10
    return True
        
for i in range(5001):
    if i == 0:
        continue
    else:
        if IsRepeat(i) == True:
            d[i] = 1
        s[i] = s[i-1] + d[i]

try:
    while True:
        N,M = input().split()
        N = int(N)
        M = int(M)
        print(s[M]-s[N]+d[N])
except Exception as e:
    pass
    

C/C++版本代碼

#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;

//#define ZANGFONG
const int maxn = 5050;
int d[maxn],s[maxn];
int a[10];
bool isrepeat(int x)
{
    int i;
    memset(a,0,sizeof(a));
    while(x)
    {
        i = x%10;
        if(a[i]==0) a[i]++;
        else return false;
        x /= 10;
    }
    return true;
}

int main()
{
    #ifdef ZANGFONG
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
    #endif // ZANGFONG
    int N,M,i;

    for(i = 1; i <= 5000; i++)
    {
        if(isrepeat(i)) d[i] = 1;
        s[i] = d[i] + s[i-1];
    }

    while(scanf("%d%d\n",&N,&M)!= EOF)
    {
        printf("%d\n",s[M]-s[N]+d[N]);
    }

    return 0;
}