You are given a string??=?1?2…??s=s1s2…sn?of length??n, which only contains digits?11,?22, ...,?99.
A substring??[?…?]s[l…r]?of??s?is a string?????+1??+2…??slsl+1sl+2…sr. A substring??[?…?]s[l…r]?of??s?is called?even?if the number represented by it is even.
Find the number of even substrings of??s. Note, that even if some substrings are equal as strings, but have different??l?and??r, they are counted as?different?substrings.
The first line contains an integer??n?(1≤?≤650001≤n≤65000)?— the length of the string??s.
The second line contains a string??s?of length??n. The string??s?consists only of digits?11,?22, ...,?99.
Print the number of even substrings of??s.
4 1234
6
4 2244
10
In the first example, the?[?,?][l,r]?pairs corresponding to even substrings are:
- ?[1…2]s[1…2]
- ?[2…2]s[2…2]
- ?[1…4]s[1…4]
- ?[2…4]s[2…4]
- ?[3…4]s[3…4]
- ?[4…4]s[4…4]
In the second example, all?1010?substrings of??s?are even substrings. Note, that while substrings??[1…1]s[1…1]?and??[2…2]s[2…2]?both define the substring "2", they are still counted as different substrings.
?
?題意:計數以偶數結尾的連續的子串的個數
?思路:遍歷一遍,如果當前數字是偶數,則表示可以以它結尾,貢獻是它所在的位置大小,加到答案里就好。
?代碼:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define ll long long
#define ull unsigned long long
#define lowbit(x) (x&(-x))
#define eps 0.00000001
#define PI acos(-1)
#define ms(x,y) memset(x, y, sizeof(x))
using namespace std;const int maxn = 65005;
char s[maxn];int main()
{int n;cin >> n;scanf("%s", s+1);ll ans = 0;for(int i=1;i<=n;i++)if((s[i] - '0') % 2 == 0)ans += i;cout << ans << endl;
}