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

题意：给你两个数m，n。问你在m～n之间0的个数。

这题类似lightoj1032，只不过这题是求0的个数。求0的个数一定要考虑是否是首位，我们知道0不能为第一位，所以这题在搜索上就加一个判断

条件，看一下搜索到的是否为首位。

#include <iostream>
#include <cstring>
using namespace std;
typedef long long LL;
LL dp[33][33];
LL dig[33];
LL dfs(LL x , LL cnt , LL flag , LL first)
{
    if(x == 0)
    {
        if(first)
            return 1;
        else
            return cnt;
    }
    if(dp[x][cnt] != -1 && !flag && !first)
        return dp[x][cnt];
    LL sum = 0;
    LL gg = flag ? dig[x] : 9;
    for(int i = 0 ; i <= gg ; i++)
    {
        if(first)
        {
            sum += dfs(x - 1 , cnt , i == gg && flag , first && i == 0);
        }
        else
        {
            if(i == 0)
            {
                sum += dfs(x - 1 , cnt + 1 , i == gg && flag , 0);
            }
            else
            {
                sum += dfs(x - 1 , cnt , i == gg && flag , 0);
            }
        }
    }
    if(!flag && !first)
        dp[x][cnt] = sum;
    return sum;
}
LL Gets(LL x)
{
    memset(dig , 0 , sizeof(dig));
    int temp = 0;
    if(x == 0)
        dig[1] = 0;
    while(x)
    {
        dig[++temp] = x % 10;
        x /= 10;
    }
    return dfs(temp , 0 , 1 , 1);
}
int main()
{
    int t;
    cin >> t;
    int ans = 0;
    while(t--)
    {
        ans++;
        LL n , m;
        memset(dp , -1 , sizeof(dp));
        cin >> n >> m;
        //cout << Gets(m) << ' ' << Gets(n - 1) << endl;
        cout << "Case " << ans << ": " << Gets(m) - Gets(n - 1) << endl;
    }
    return 0;
}