hdu4722之简单数位dp

 

Good Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 428    Accepted Submission(s): 149

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 <= 10 ¹⁸).

 

 

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

2013 ACM/ICPC Asia Regional Online —— Warmup2

#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];//分别代表长度为i位数和mod 10为j的个数 

int digit[MAX];



void digit_dp(){//计算每长度为i为的数mod 10 == 0的个数 

	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][((0-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;

}