Codeforce Round #213 Div2 C

You have a string of decimal digits s. Let's define b_ij = s_i·s_j. Find in matrix b the number of such rectangles that the sum b_ij for all cells (i, j) that are the elements of the rectangle equals a in each rectangle.

A rectangle in a matrix is a group of four integers (x, y, z, t) (x ≤ y, z ≤ t). The elements of the rectangle are all cells (i, j) such that x ≤ i ≤ y, z ≤ j ≤ t.

Input

The first line contains integer a (0 ≤ a ≤ 10⁹), the second line contains a string of decimal integers s (1 ≤ |s| ≤ 4000).

Output

Print a single integer — the answer to a problem.

Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Sample test(s)

Input

10 
12345

Output

Input

16 
439873893693495623498263984765

Output

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <vector>
 4 #include <set>
 5 #include <cstring>
 6 #include <string>
 7 #include <map>
 8 #include <cmath>
 9 #include <ctime>
10 #include <algorithm>
11 #include <queue>
12 
13 using namespace std;
14 
15 #define mp make_pair
16 #define pb push_back
17 #define rep(i,n) for(int i = 0; i < (n); i++)
18 #define re return
19 #define fi first
20 #define se second
21 #define sz(x) ((int) (x).size())
22 #define all(x) (x).begin(), (x).end()
23 #define sqr(x) ((x) * (x))
24 #define sqrt(x) sqrt(abs(x))
25 #define y0 y3487465
26 #define y1 y8687969
27 #define fill(x,y) memset(x,y,sizeof(x))
28 
29 typedef vector<int> vi;
30 typedef long long ll;
31 typedef long double ld;
32 typedef double D;
33 typedef pair<int, int> ii;
34 typedef vector<ii> vii;
35 typedef vector<string> vs;
36 typedef vector<vi> vvi;
37 
38 template<class T> T abs(T x) { re x > 0 ? x : -x; }
39 
40 const int N = 40000;
41 
42 int n;
43 int m;
44 //string s;
45 
46 int cnt[N];
47 char s[4010];
48 int sum[4010];
49 int main(){
50     scanf("%d%s", &n, &s);
51     m = strlen(s);
52     sum[0] = 0;
53     for (int i = 1; i <= m; i++){ sum[i] = sum[i - 1] + s[i - 1] - '0'; }
54     for (int i = 0; i < m; i++){
55         int cur = 0;
56         for (int j = i; j < m; j++){
57             cur += s[j] - '0';
58             cnt[cur]++;
59         }
60     }
61     ll ans = 0;
62     for (int i = 1; i <= m; i++)
63     for (int j = i; j <= m; j++){
64         if (sum[j] - sum[i - 1] != 0 && n % (sum[j] - sum[i - 1]) == 0 && n / (sum[j] - sum[i - 1])<N)
65             ans += cnt[n / (sum[j] - sum[i - 1])];
66         else if (n == 0 && sum[j] - sum[i - 1] == 0)ans += (ll) m*(m + 1) / 2;
67     }
68     printf("%I64d\n", ans);
69     return 0;
70 }

View Code

posted @ 2013-11-20 12:09 HaibaraAi 阅读(164) 评论(0) 收藏举报

刷新页面返回顶部