HDOJ 2089 不要62

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题意：求m到n中不含4与62(连续)的数的个数

链接：http://acm.hdu.edu.cn/showproblem.php?pid=2089

思路：数位DP(不会)或树状数组。树状数组：预处理，单点更新，区间查询

注意点：无

以下为AC代码：

Run IDSubmit TimeJudge StatusPro.IDExe.TimeExe.MemoryCode Len.LanguageAuthor129410052015-02-12 20:09:15Accepted208993MS4508K1524 BG++luminous11

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <vector>#include <deque>#include <list>#include <cctype>#include <algorithm>#include <climits>#include <queue>#include <stack>#include <cmath>#include <map>#include <set>#include <iomanip>#include <cstdlib>#include <ctime>#define ll long long#define ull unsigned long long#define all(x) (x).begin(), (x).end()#define clr(a, v) memset( a , v , sizeof(a) )#define pb push_back#define mp make_pair#define read(f) freopen(f, "r", stdin)#define write(f) freopen(f, "w", stdout)using namespace std;//const double pi = acos(-1);//const double eps = 1e-10;//const int dir[[4][2] = { 1,0, -1,0, 0,1, 0,-1 };int sum[1000005] = { 0 };bool calc ( int num ){    while ( num ){        if ( ( num % 100 == 62 || num % 10 == 4 ) )            return false;        else            num /= 10;    }    return true;}void add ( int k, int num ){    for ( ; k < 1000005; k += ( k & ( -k ) ) )        sum[k] += num;}int query ( int k ){    int ans = 0;    for ( ; k > 0; k -= ( k & ( -k ) ) )        ans += sum[k];    return ans;}int main(){    ios::sync_with_stdio( false );    int m, n;    for ( int i = 1; i < 1000005; i ++ ){        if ( calc ( i ) ){            add ( i, 1 );        }    }    while ( cin >> m >> n ){        if ( m == 0 && n == 0 )break;        cout << query ( n ) - query ( m - 1 ) << endl;    }    return 0;}

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