lightoj 1140 - 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

Output for Sample Input

5

10 11

100 200

0 500

1234567890 2345678901

0 4294967295

Case 1: 1

Case 2: 22

Case 3: 92

Case 4: 987654304

Case 5: 3825876150

给你一个区间，问这个区间里所有数字共有多少个0。

注意前导零的判断，还有得开二维的，一维的不知道怎么写。。。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define inf 1e9#define endl '\n'#define LL long longusing namespace std;LL dp[20][20];int a[20];LL dfs(int len,int s,int num,int fp)//s判断有没有前导零{    if(!len)        return s == 0 ? (LL)1 : (LL)num;    if(!fp && s && dp[len][num] != -1)        return dp[len][num];    int n = fp ? a[len]:9;    LL res = 0;    for(int i=0;i<=n;i++)    {        if(i==0)        {            if(s)                res += dfs(len-1,s,num + 1,fp&&i==n);            else                res += dfs(len-1,0,num,fp&&i==n);        }        else            res += dfs(len-1,1,num,fp&&i==n);    }    if(!fp && s)        dp[len][num] = res;    return res;}LL sum(LL x){    int len = 0;    while(x)    {        a[++len] = x % 10;        x /= 10;    }    return dfs(len,0,0,1);}int main(void){    int T;    scanf("%d",&T);    memset(dp,-1,sizeof(dp));    int cas = 1;    while(T--)    {        LL n,m;        scanf("%lld%lld",&n,&m);        LL ans = sum(m) - sum(n-1);        printf("Case %d: %lld\n",cas++,ans);    }    return 0;}