1140 - How Many Zeroes?

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Time Limit: 2 second(s)

Memory Limit: 32 MB

Jimmy writes down the decimal representations of all natural numbers between and including m and n, (m ≤ n). How many zeroes will he write down?

Input

Input starts with an integer T (≤ 11000), denoting the number of test cases.

Each case contains two unsigned 32-bit integers m and n, (m ≤ n).

Output

For each case, print the case number and the number of zeroes written down by Jimmy.

Sample Input	Output for Sample Input
5 10 11 100 200 0 500 1234567890 2345678901 0 4294967295	Case 1: 1 Case 2: 22 Case 3: 92 Case 4: 987654304 Case 5: 3825876150

 

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SPECIAL THANKS: JANE ALAM JAN (DESCRIPTION, SOLUTION, DATASET)

简单数位DP

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int a[30];
ll dp[30][30];
ll dfs(int pos,int sta,int lead,int limit)
{
	if(pos==-1){
		return sta;
	}
	if(!limit&&!lead&&dp[pos][sta]!=-1) 
	return dp[pos][sta];
	int up=limit?a[pos]:9;
	ll tmp=0;
	for(int i=0;i<=up;i++)
	{

	if(lead==1&&i==0)     //前导0
	{
	tmp+=dfs(pos-1,sta,lead,limit&&i==up);
	}
	else 
	{ 
	if(i==0)                     
	tmp+=dfs(pos-1,sta+1,lead && i==0,limit&&i==up);
	else
	tmp+=dfs(pos-1,sta,lead && i==0,limit&&i==up);
	}
	}
	if(!limit&&!lead) 
	dp[pos][sta]=tmp;
	return tmp; 
}
 
ll solve(ll x)
{
    int pos=0;
    while(x)
    {
        a[pos++]=x%10;
        x/=10;
    }
    return dfs(pos-1,0,true,true);
}
int main()
{
	
    memset(dp,-1,sizeof(dp));
    int T;
    cin>>T;
    for(int i=1;i<=T;i++)
    {
	ll n,m;
	scanf("%lld%lld",&n,&m);
	if(n==0)
	printf("Case %d: %lld\n",i,solve(m)-solve(n-1)+1);//0是一个特殊情况
	else
	printf("Case %d: %lld\n",i,solve(m)-solve(n-1));
    }

    return 0;
}