AtCoder Beginner Contest 124 D - Handstand（思维+前缀和）

D - Handstand

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Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 400400 points

Problem Statement

NN people are arranged in a row from left to right.

You are given a string SS of length NN consisting of 0 and 1, and a positive integer KK.

The ii-th person from the left is standing on feet if the ii-th character of SS is 0, and standing on hands if that character is 1.

You will give the following direction at most KK times (possibly zero):

Direction: Choose integers ll and rr satisfying 1≤l≤r≤N1≤l≤r≤N, and flip the ll-th, (l+1)(l+1)-th, ......, and rr-th persons. That is, for each i=l,l+1,...,ri=l,l+1,...,r, the ii-th person from the left now stands on hands if he/she was standing on feet, and stands on feet if he/she was standing on hands.

Find the maximum possible number of consecutive people standing on hands after at most KK directions.

Constraints

• NN is an integer satisfying 1≤N≤1051≤N≤105.
• KK is an integer satisfying 1≤K≤1051≤K≤105.
• The length of the string SS is NN.
• Each character of the string SS is 0 or 1.

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Input

Input is given from Standard Input in the following format:

NN KK
SS

Output

Print the maximum possible number of consecutive people standing on hands after at most KK directions.

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Sample Input 1 Copy

Copy

5 1
00010

Sample Output 1 Copy

Copy

4

We can have four consecutive people standing on hands, which is the maximum result, by giving the following direction:

• Give the direction with l=1,r=3l=1,r=3, which flips the first, second and third persons from the left.

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Sample Input 2 Copy

Copy

14 2
11101010110011

Sample Output 2 Copy

Copy

8

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Sample Input 3 Copy

Copy

1 1
1

Sample Output 3 Copy

Copy

1

No directions are necessary.

题意：

给你一个只含有0和1的字符串，并且给你一个数字K，你可以选择最多K次区间，每一个区间L<=R，然后对这个区间的元素进行取反操作。

即0变成1，1变成0，问你在最聪明的操作之后最大可以获得的连续1的串是多长？

可以看样例解释理解题意。

思路：

我们对字符串的连续1和0串进行计数处理，即把连续的1或者0的个数记录起来，

那么字符串会生成如下的数组

例如字符串是  110000111001101

我们把连续的1和连续0分别放入a和b数组，

那么a的元素是2 3 2 1

b的元素是4 2 1

我们思考可以发现，我们想要最长的连续1串，那么我们处理的区间肯定是连续的0区间，让他们变成1，然后来增长连续1串的长度。

即处理的0区间一定是连续的。

那么我们可以知道如下，例如我们处理两个区间，那么可以的得到的最长的连续1串就是这两个区间的0串长度的sum和以及这两个0串前后和中间的1串的sum和。

那么我们只需要枚举连续的K个的0串即可，

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=;while(b){if(b%)ans=ans*a%MOD;a=a*a%MOD;b/=;}return ans;}
inline void getInt(int* p);
const int maxn=;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int n;
int k;
char s[maxn];
std::vector<int> v1,v2;
ll sum1[maxn];
ll sum2[maxn];
int main()
{
    //freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
    //freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);
    gbtb;
    cin>>n>>k;
    cin>>s;
    int cnt=;
//    int i=0;
    if(s[]=='')
    {
        v1.pb();
    }
    repd(i,,n-)
    {
        if(s[i]=='')
        {
            while(s[i]=='')
            {
                cnt++;
                i++;
            }
            v2.push_back(cnt);
            cnt=;
            i--;
        }
        if(s[i]=='')
        {
            while(s[i]=='')
            {
                cnt++;
                i++;
            }
            v1.push_back(cnt);
            cnt=;
            i--;
        }

    }
    int num=max(sz(v1),sz(v2));
    repd(i,,maxn-)
    {
        v1.push_back();
        v2.push_back();
    }
    repd(i,,maxn-)
    {
        sum1[i]+=sum1[i-]+v1[i-];
        sum2[i]+=sum2[i-]+v2[i-];
    }
    int w=;
    ll ans=0ll;
    repd(i,,n-k+)
    {
        int l=i;
        int r=l+k;
        ll temp=sum1[r]-sum1[l-];
        temp+=sum2[r-]-sum2[l-];
        ans=max(ans,temp);
    }
    cout<<ans<<endl;

    return ;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '');
        while ((ch = getchar()) >= '' && ch <= '') {
            *p = *p *  - ch + '';
        }
    }
    else {
        *p = ch - '';
        while ((ch = getchar()) >= '' && ch <= '') {
            *p = *p *  + ch - '';
        }
    }
}