【HDU】3555 Bomb 数位DP

传送门：【HDU】3555 Bomb

题目分析：数位DP水题。

代码如下：

#include <cstdio>#include <cstring>#include <algorithm>//#include <cmath>#include <queue>using namespace std ;  typedef long long LL ;  #define rep( i , a , b ) for ( int i = a ; i < b ; ++ i )#define For( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define rev( i , a , b ) for ( int i = a ; i >= b ; -- i )#define travel( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next )#define clr( a , x ) memset ( a , x , sizeof a )#define cpy( a , x ) memcpy ( a , x , sizeof a )const int MAXN = 65 ;LL dp[MAXN][10][2] ;LL ten[MAXN] ;int vis[MAXN][10][2] , Time ;int num[MAXN] ;LL n ;LL dfs ( int cur , int j , int flag , int num[] ) {if ( cur == -1 ) return 0 ;if ( vis[cur][j][flag] == Time ) return dp[cur][j][flag] ;vis[cur][j][flag] = Time ;dp[cur][j][flag] = 0 ;if ( flag == 0 ) {rep ( i , 0 , num[cur] ) {if ( j == 4 && i == 9 ) dp[cur][j][flag] += ten[cur] ;else dp[cur][j][flag] += dfs ( cur - 1 , i , 1 , num ) ;}if ( j == 4 && num[cur] == 9 ) dp[cur][j][flag] += n % ten[cur] + 1 ;else dp[cur][j][flag] += dfs ( cur - 1 , num[cur] , 0 , num ) ;} else {rep ( i , 0 , 10 ) {if ( j == 4 && i == 9 ) dp[cur][j][flag] += ten[cur] ;else dp[cur][j][flag] += dfs ( cur - 1 , i , 1 , num ) ;}}return dp[cur][j][flag] ;}void solve () {LL tmp ;int n1 = 0 ;scanf ( "%I64d" , &n ) ;tmp = n ;while ( tmp ) {num[n1 ++] = tmp % 10 ;tmp /= 10 ;}++ Time ;LL count = dfs ( n1 - 1 , 0 , 0 , num ) ;printf ( "%I64d\n" , count ) ;}int main () {//freopen ( "in.txt" , "r" , stdin ) ;//freopen ( "out.txt" , "w" , stdout ) ;int T ;clr ( vis , 0 ) ;Time = 0 ;ten[0] = 1 ;rep ( i , 1 , MAXN ) ten[i] = ten[i - 1] * 10 ;scanf ( "%d" , &T ) ;while ( T -- ) solve () ;return 0 ;}