lightoj1140_数位dp

题目链接：http://lightoj.com/volume_showproblem.php?problem=1140

给出区间[m,n]，求区间内的所有数共有多少个0。

设dp[i][j]表示处理到第i位时，它前面共有j个0（除了前导零）。

注意：前导0

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <cstdio>
 6 #include <vector>
 7 #include <ctime>
 8 #include <queue>
 9 #include <list>
10 #include <set>
11 #include <map>
12 using namespace std;
13 #define INF 0x3f3f3f3f
14 typedef long long LL;
15 
16 LL dp[12][12];
17 int bit[12];
18 LL dfs(int len, int zer, int sum, int flag)
19 {
20     if(len < 1){
21         if(zer == 0)
22             return 1;
23         return sum;
24     }
25     if(dp[len][sum] != -1 && flag && zer)//注意zer
26         return dp[len][sum];
27     int te = flag ? 9 : bit[len];
28     LL res = 0;
29     for(int i = 0; i <= te; i++)
30     {
31         int Sum = sum, Zer = zer;
32         if(zer == 0 && i != 0)
33             Zer = 1;
34         if(i == 0 && zer == 1)
35             Sum++;
36         res += dfs(len-1, Zer, Sum, flag || i < te);
37     }
38     if(flag && zer)//注意zer,一定要写
39         dp[len][sum] = res;
40     return res;
41 }
42 LL solve(LL n)
43 {
44     int len = 0;
45     memset(bit, 0, sizeof(bit));
46     while(n)
47     {
48         bit[++len] = n % 10;
49         n /= 10;
50     }
51     return dfs(len, 0, 0, 0);
52 }
53 int main()
54 {
55     int t;
56     LL l, r;
57     memset(dp, -1, sizeof(dp));
58     scanf("%d", &t);
59     for(int ca = 1; ca <= t; ca++)
60     {
61         scanf("%lld %lld", &l, &r);
62         printf("Case %d: %lld
", ca, solve(r)-solve(l-1));
63     }
64     return 0;
65 }