Lightoj1140——How Many Zeroes?（数位dp）

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

Output for Sample Input

Case 1: 1

Case 2: 22

Case 3: 92

Case 4: 987654304

Case 5: 3825876150

求m到n的数共有多少个0

要注意会出现01,001,0001这些前导0的情况，用first判断首位是否是0

lightoj的64位输入输出差点把我又坑一次

#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <cstdio>
#include <set>
#include <cmath>
#include <map>
#include <algorithm>
#define INF 0x3f3f3f3f
#define MAXN 10000005
#define Mod 10001
using namespace std;
int dight[30];
long long dp[20][20];
long long dfs(int pos,int s,bool limit,int first)
{
if(pos==0)
{
if(first)
return 1;
return s;
}
if(!limit&&dp[pos][s]!=-1&&!first)
return dp[pos][s];
int end;
long long ret=0;
if(limit)
end=dight[pos];
else
end=9;
for(int d=0; d<=end; ++d)
{
if(first)
ret+=dfs(pos-1,0,limit&&d==end,first&&d==0);
else
{
if(d==0)
ret+=dfs(pos-1,s+1,limit&&d==end,0);
else
ret+=dfs(pos-1,s,limit&&d==end,0);
}
}
if(!limit&&!first)
dp[pos][s]=ret;
return ret;

}
long long solve(long long a)
{
memset(dight,0,sizeof(dight));
int cnt=1;
while(a!=0)
{
dight[cnt++]=a%10;
a/=10;
}
return dfs(cnt-1,0,1,1);
}
int main()
{
memset(dp,-1,sizeof(dp));
int t,cnt=1;
scanf("%d",&t);
while(t--)
{
long long x,y;
scanf("%lld%lld",&x,&y);
printf("Case %d: %lld\n",cnt++,solve(y)-solve(x-1));
}
return 0;
}