How Many Zeroes? LightOJ - 1140

题目：

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

大意：

给你一个范围，求这个范围内的数字中出现0的次数

思路：

数位dp，注意按位dp时出现前导0

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 1000000000;
const int maxn = 123456;
int T;
ll l,r,ans;
ll a[25];
ll dp[25][25];//dp[i][j]处理到第i位时前面出现了j个0

ll dfs(int pos,bool limt,int cnt,bool is0)//is0表示前面是否一直为0
{
if(!pos)//计算到最后一位
return is0?1:cnt;//前导0之类的算一个0
ll sum=0;
if(!limt&&dp[pos][cnt]!=-1&&!is0)
return dp[pos][cnt];
int k=(limt?a[pos]:9);
for(int i=0;i<=k;i++)
{
if(is0)
sum+=dfs(pos-1,limt&&(i==k),0,i==0);
else
sum+=dfs(pos-1,limt&&(i==k),cnt+(i==0),0);
}
if(!limt&&!is0)
dp[pos][cnt]=sum;
return sum;
}

ll solve(ll k)
{
int p=0;
while(k)
{
a[++p]=k%10;
k/=10;
}
return dfs(p,1,0,1);
}

int main()
{
int no=0;
scanf("%d",&T);
while(T--)
{
scanf("%lld%lld",&l,&r);
memset(dp,-1,sizeof(dp));
printf("Case %d: ",++no);
printf("%lld\n",solve(r)-solve(l-1));
}
return 0;
}