LightOJ

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

5

10 11

100 200

0 500

1234567890 2345678901

0 4294967295

Sample Output

Case 1: 1

Case 2: 22

Case 3: 92

Case 4: 987654304

Case 5: 3825876150

题意：从l到r中，所有数字中含有的0的个数。不包括前导0

解题思路：数位DP，注意判断前导0.

#include<iostream>#include<deque>#include<memory.h>#include<stdio.h>#include<map>#include<string.h>#include<algorithm>#include<vector>#include<math.h>#include<stack>#include<queue>#include<set>using namespace std;typedef long long int ll;ll dp[30][30];//遍历到第i位，前面有j个0的0的个数。int dig[30];//iszero记录是否有前导0ll dfs(int pos,int zeronum,bool limit,bool iszero){        if(pos==-1){        if(iszero)//如果有前导0，且到了最后一位，返回1,0也算一个0            return 1;        else            return zeronum;    }            //记忆化搜索    if(!limit && dp[pos][zeronum]!=-1&&!iszero)        return dp[pos][zeronum];            //循环遍历下一位的每一个数字，end记录最大能达到的数    int end=limit?dig[pos]:9;    ll ans=0;        for(int i=0;i<=end;i++){                if(iszero)            ans+=dfs(pos-1,0,limit&&i==end,i==0);//当前还是不是0        else{            if(i==0)                ans+=dfs(pos-1,zeronum+1,limit&&i==end,0);//非前导零的0要把0数加1            else                ans+=dfs(pos-1,zeronum,limit&&i==end,0);        }            }            if(!limit&&!iszero)        dp[pos][zeronum]=ans;        return ans;    }ll solve(ll x){    //记录每一位    int pos=0;    while(x){        dig[pos++]=x%10;        x/=10;    }        return dfs(pos-1,0,1,1);}int main(){    ll l,r;    int t;    scanf("%d",&t);    for(int qqq=1;qqq<=t;qqq++){        scanf("%lld%lld",&l,&r);        memset(dp,-1,sizeof(dp));        printf("Case %d: %lld\n",qqq,solve(r)-solve(l-1));            }        return 0;}