Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 622 Accepted Submission(s): 380
Problem Description A DFS(digital factorial sum) number is found by summing the factorial of every digit of a positive integer. For example ,consider the positive integer 145 = 1!+4!+5!, so it's a DFS number. Now you should find out all the DFS numbers in the range of int( [1, 2147483647] ). There is no input for this problem. Output all the DFS numbers in increasing order. The first 2 lines of the output are shown below. Input no input Output Output all the DFS number in increasing order. Sample Output 1 2 ...... Author zjt Recommend lcy Statistic | Submit | Back // 1364495 2009-05-13 20:42:44 Time Limit Exceeded 2212 2000MS 232K 426 B C++ Wpl // 1364655 2009-05-13 21:03:44 Accepted 2212 0MS 204K 299 B C++ Wpl /* For example ,consider the positive integer 145 = 1!+4!+5!, so it's a DFS number. Now you should find out all the DFS numbers in the range of int( [1, 2147483647] ). */ #include < iostream > #define MAX 5 using namespace std; int data[MAX]; int main() { data[ 0 ] = 1 ; data[ 1 ] = 2 ; data[ 2 ] = 145 ; data[ 3 ] = 40585 ; int i; for (i = 0 ;i <= 3 ;i ++ ) printf( " %d\n " ,data[i]); return 0 ; }附上打表的程序
// 1364495 2009-05-13 20:42:44 Time Limit Exceeded 2212 2000MS 232K 426 B C++ Wpl #include < iostream > #include < fstream > #define MAX 10000 using namespace std; int f[ 11 ],data[MAX]; bool DFS( int n) { int sum = 0 ,x = n; while (x != 0 ) { sum += f[x % 10 ]; x = x / 10 ; if (sum > n) return false ; } if (sum == n) return true ; else return false ; } int main() { int i,j; f[ 0 ] = 1 ; for (i = 1 ;i <= 10 ;i ++ ) f[i] = i * f[i - 1 ]; ofstream outfile( " ans.txt " ); j = 0 ; for (i = 1 ;i < 2147483647 ;i ++ ) { if (DFS(i)) { outfile << " data[ " << j ++<< " ]= " << i << endl; } } return 0 ; }
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