[暑假集训--数位dp]LightOJ1140 How Many Zeroes?

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

问 l 到 r 出现了几个0

记一下出现了几个0，有没有前导零就好

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<algorithm>
 6 #include<cmath>
 7 #include<queue>
 8 #include<deque>
 9 #include<set>
10 #include<map>
11 #include<ctime>
12 #define LL long long
13 #define inf 0x7ffffff
14 #define pa pair<int,int>
15 #define mkp(a,b) make_pair(a,b)
16 #define pi 3.1415926535897932384626433832795028841971
17 using namespace std;
18 inline LL read()
19 {
20     LL x=0,f=1;char ch=getchar();
21     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
22     while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
23     return x*f;
24 }
25 LL n,len,l,r;
26 LL f[20][20][20][2];
27 int d[110];
28 int zhan[20],top;
29 inline LL dfs(int now,int dat,int tot,int lead,int fp)
30 {
31     if (now==1)return tot;
32     if (!fp&&f[now][dat][tot][lead]!=-1)return f[now][dat][tot][lead];
33     LL ans=0,mx=fp?d[now-1]:9;
34     for (int i=0;i<=mx;i++)
35     {
36         int nexlead=lead&&i==0&&now-1!=1;
37         ans+=dfs(now-1,i,tot+(i==0&&!nexlead),nexlead,fp&&i==d[now-1]);
38     }
39     if (!fp)f[now][dat][tot][lead]=ans;
40     return ans;
41 }
42 inline LL calc(LL x)
43 {
44     if (x<=0)return x+1;
45     LL xxx=x;
46     len=0;
47     while (xxx)
48     {
49         d[++len]=xxx%10;
50         xxx/=10;
51     }
52     LL sum=0;
53     for (int i=0;i<=d[len];i++)
54     {
55         if (i)sum+=dfs(len,i,0,0,i==d[len]);
56         else sum+=dfs(len,0,len==1,1,!d[len]);
57     }
58     return sum;
59 }
60 main()
61 {
62     int T=read(),cnt=0;
63     while (T--)
64     {
65         l=read();r=read();
66         if (r<l)swap(l,r);
67         memset(f,-1,sizeof(f));
68         printf("Case %d: %lld\n",++cnt,calc(r)-calc(l-1));
69     }
70 }

LightOJ 1140