CodeForces 165C Another Problem on Strings（公式推导）

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 string w. 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, cout streams or the %I64d specifier.

Example

Input

11010

Output

6

Input

201010

Output

4

Input

10001010

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".

题解：

题意：

找串中满足有n个1的所有子串个数，注意位置不一样长得一样也是不一样的子串，当n为0时要特殊处理

找规律发现n为0的时候答案为连续0的长度*（长度+1）/2的累加

不为0时找左右1的位置，从而算出左右0的个数，相乘累加

这题用%lld会报错要用%I64d

代码：

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<queue>#include<stack>#include<math.h>#include<vector>#include<map>#include<set>#include<stdlib.h>#include<cmath>#include<string>#include<algorithm>#include<iostream>#include<stdio.h>using namespace std;#define ll long longchar s[1000005];int a[1000005];int main(){    int i,j,k,n,len=0,x,d;    scanf("%d",&n);    scanf("%s",s);    d=strlen(s);    for(i=0;s[i];i++)    {        if(s[i]=='1')        {            a[len]=i;            len++;        }    }    if(len<n)    {        printf("0\n");        return 0;    }    if(n==0)    {        if(len==d)        {            printf("0\n");            return 0;        }        ll ans=0,add=0;        for(i=0;s[i];i++)        {            if(s[i]=='0')                add++;            else            {                if(add!=0)                    ans+=(add*(add+1)/2);                add=0;            }        }        if(add!=0)            ans+=(add*(add+1)/2);        printf("%I64d\n",ans);        return 0;    }    int ed;    ll r,l,ans=0;    for(i=0;;i++)    {        ed=i+n;        if(ed>len)            break;        l=1,r=1;        if(i!=0)        {            l+=(a[i]-a[i-1]-1);        }        else        {            l+=a[i];        }        if(ed!=len)        {            r+=(a[ed]-a[ed-1]-1);        }        else        {            r+=(d-a[ed-1]-1);        }        ans+=l*r;    }    printf("%I64d\n",ans);    return 0;}