hdu4722 Good Numbers(數位dp)
阿新 • • 發佈:2018-12-23
Good Numbers
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3529 Accepted Submission(s): 1128
Problem Description If we sum up every digit of a number and the result can be exactly divided by 10, we say this number is a good number.
You are required to count the number of good numbers in the range from A to B, inclusive.
Input The first line has a number T (T <= 10000) , indicating the number of test cases.
Each test case comes with a single line with two numbers A and B (0 <= A <= B <= 1018
Output For test case X, output "Case #X: " first, then output the number of good numbers in a single line.
Sample Input 2 1 10 1 20
Sample Output Case #1: 0 Case #2: 1 Hint The answer maybe very large, we recommend you to use long long instead of int.
Source 題意:求a到b(並不包括b)之間共有多少個數能被10整除。 分析:基礎題,dp[ i ][ j ]表示長度為 i 的數對10取模的值為 j 。
#include <iostream> #include <cstdio> #include <cstring> #include <stack> #include <queue> #include <map> #include <set> #include <vector> #include <cmath> #include <algorithm> using namespace std; const double eps = 1e-6; const double pi = acos(-1.0); const int INF = 0x3f3f3f3f; const int MOD = 1000000007; #define ll long long #define CL(a) memset(a,0,sizeof(a)) ll s[20]; ll dp[20][10];//dp[i][j]表示長度為i的數對10取模為j ll slove(ll x) { ll t=0,sum=0; while (x) { s[++t]=x%10; x/=10; } ll ans=0,m=0; CL(dp); for (int i=t; i>0; i--)//最高位開始列舉 { for (int j=0; j<10; j++)//沒有界限,列舉所有 for (int k=0; k<10; k++) dp[i][(j+k)%10]+=dp[i+1][j]; for (int j=0; j<s[i]; j++)//有界限,如上一位為1,該位為2;而上一位已經是1了,所以該位只能取到2 dp[i][(j+m)%10]++; m = (m+s[i])%10;//儲存餘數 } if (!m) dp[1][0]++; return dp[1][0]; } int main () { int T,ii=1; ll a,b; scanf ("%d",&T); while (T--) { scanf ("%lld%lld",&a,&b); printf ("Case #%d: %lld\n",ii++,slove(b)-slove(a-1)); } return 0; }