hdu4722之简单数位dp

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


Problem DescriptionIf 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.
InputThe 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).
OutputFor test case X, output "Case #X: " first, then output the number of good numbers in a single line.
Sample Input

21 101 20

Sample Output

Case #1: 0Case #2: 1HintThe answer maybe very large, we recommend you to use long long instead of int. 

Source2013 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 99999999using 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;}