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
",++num,calculate(b+1)-calculate(a));
	}
	return 0;
}