How Many Zeroes? LightOJ

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0ll 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;}