Codeforces Round #112 (Div. 2) C. Another Problem on Strings

C. Another Problem on Strings

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A string is binary, if it consists only of characters "0" and "1".

String v is a substring of string w if it has a non-zero length and can be read starting from some position in stringw. For example, string "010" has six substrings: "0", "1", "0", "01", "10", "010". Two substrings are considered different if their positions of occurrence are different. So, if some string occurs multiple times, we should consider it the number of times it occurs.

You are given a binary string s. Your task is to find the number of its substrings, containing exactly k characters "1".

Input

The first line contains the single integer k (0 ≤ k ≤ 106). The second line contains a non-empty binary string s. The length of s does not exceed 106 characters.

Output

Print the single number — the number of substrings of the given string, containing exactly k characters "1".

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

Sample test(s)

input

1
1010

output

6

input

2
01010

output

4

input

100
01010

output

0

Note

In the first sample the sought substrings are: "1", "1", "10", "01", "10", "010".

In the second sample the sought substrings are: "101", "0101", "1010", "01010".

題意：給你一個01字符串，讓你求它連續子串中滿足有k個一的子串個數。

思路：就是統計前i個字符的和爲k的個數，當sum值大於等於k時，就可以將子串起點後移相應位數，用dp[sum]表示前i個字符和值爲sum的個數

代碼：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 1000005
#define LL long long

char s[N];
int dp[N];//記錄前i個字符中組成各個和值的個數
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        scanf("%s",s);
        int i;
        int len=strlen(s);
        int sum=0;
        dp[0]=1;
        LL ans=0;
        for(i=0;i<len;i++)
        {
            if(s[i]=='1')
            {
                sum++;
            }
            if(sum-n>=0)
            ans+=dp[sum-n]*1;//在當前sum值下，和值爲n的數增加的個數
            dp[sum]++;//每統計一次和爲sum的就增加1
        }
        printf("%lld\n",ans);
    }
    return 0;
}