HDU 4722 數位DP

Good Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3620    Accepted Submission(s): 1147

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

#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<queue> #include<algorithm> #include<map> #include<iomanip> #define INF 99999999 using namespace std; const int MAX=22; __int64 dp[MAX][10];//分別代表長度爲i位數和mod 10爲j的個數  int digit[MAX]; void digit_dp(){//計算每長度爲i爲的數mod 10 == 0的個數  dp[0][0]=1; for(int i=1;i<MAX;++i){ for(int j=0;j<10;++j){ for(int k=0;k<10;++k){ dp[i][j]+=dp[i-1][(j-k+10)%10]; } } } } __int64 calculate(__int64 n){ int size=0,last=0; __int64 sum=0; while(n)digit[++size]=n%10,n/=10; for(int i=size;i>=1;--i){ for(int j=0;j<digit[i];++j){ sum+=dp[i-1][((j+last)%10+10)%10]; } last=(last+digit[i])%10; } return sum; } int main(){ digit_dp(); int t,num=0; __int64 a,b; scanf("%d",&t); while(t--){ scanf("%I64d%I64d",&a,&b); printf("Case #%d: %I64d\n",++num,calculate(b+1)-calculate(a)); } return 0; }