【HDU4722】【Good Numbers】

Good Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3470    Accepted Submission(s): 1099

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

21 101 20

 

Sample Output

Case #1: 0Case #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

 

Recommend

zhuyuanchen520

<span style="font-size:14px;background-color: rgb(255, 255, 153);">return dp[len][n] = ret;</span>

这句话开始写错了，然后各种wa

#include <iostream>#include <cstring>#include <cmath>#include <queue>#include <stack>#include <list>#include <map>#include <set>#include <string>#include <cstdlib>#include <cstdio>#include <algorithm>using namespace std;    __int64 T;__int64 dp[20][25];char s[50];__int64 f(int len,int n) // return a single ans 长度为len,各位和mod10 = n{if(len == 0) return n == 0;if(dp[len][n]>=0 ) return dp[len][n];dp[len][n] = 0;__int64 ret = 0LL;for(int i=9;i>=0;i--){ret += f(len-1,i);}return dp[len][n] = ret;}__int64 solve(__int64 x) // return all the ans of <=x {__int64 ret = 0;sprintf(s,"%I64d",x);int len = strlen(s);int curlen = len;int sum = 0;for(int i=0;i<len;i++){int d = s[i] - '0';curlen --;for(int j=0;j<d;j++){int tmp = (sum + j) % 10; ret += f(curlen,(10-tmp) % 10);}sum = (sum + d) % 10;}return ret;}int main(){    scanf("%I64d",&T);int C = 1;memset(dp,-1,sizeof(dp));while(T--){__int64 l,r;scanf("%I64d%I64d",&l,&r);printf("Case #%d: %I64d\n",C++,solve(r+1)-solve(l));cout << endl;}    return 0;}