[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

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #include<deque>
 #include<set>
 #include<map>
 #include<ctime>
 #define LL long long
 #define inf 0x7ffffff
 #define pa pair<int,int>
 #define mkp(a,b) make_pair(a,b)
 #define pi 3.1415926535897932384626433832795028841971
 using namespace std;
 inline LL read()
 {
     LL x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 LL n,len,l,r;
 LL f[][][][];
 int d[];
 int zhan[],top;
 inline LL dfs(int now,int dat,int tot,int lead,int fp)
 {
     if (now==)return tot;
     if (!fp&&f[now][dat][tot][lead]!=-)return f[now][dat][tot][lead];
     LL ans=,mx=fp?d[now-]:;
     for (int i=;i<=mx;i++)
     {
         int nexlead=lead&&i==&&now-!=;
         ans+=dfs(now-,i,tot+(i==&&!nexlead),nexlead,fp&&i==d[now-]);
     }
     if (!fp)f[now][dat][tot][lead]=ans;
     return ans;
 }
 inline LL calc(LL x)
 {
     if (x<=)return x+;
     LL xxx=x;
     len=;
     while (xxx)
     {
         d[++len]=xxx%;
         xxx/=;
     }
     LL sum=;
     for (int i=;i<=d[len];i++)
     {
         if (i)sum+=dfs(len,i,,,i==d[len]);
         else sum+=dfs(len,,len==,,!d[len]);
     }
     return sum;
 }
 main()
 {
     int T=read(),cnt=;
     while (T--)
     {
         l=read();r=read();
         if (r<l)swap(l,r);
         memset(f,-,sizeof(f));
         printf("Case %d: %lld\n",++cnt,calc(r)-calc(l-));
     }
 }

LightOJ 1140