You are given a garland consisting of n lamps. States of the lamps are represented by the string s of length n. The i-th character of the string si equals ‘0’ if the i-th lamp is turned off or ‘1’ if the i-th lamp is turned on. You are also given a positive integer k.

In one move, you can choose one lamp and change its state (i.e. turn it on if it is turned off and vice versa).

The garland is called k-periodic if the distance between each pair of adjacent turned on lamps is exactly k. Consider the case k=3. Then garlands “00010010”, “1001001”, “00010” and “0” are good but garlands “00101001”, “1000001” and “01001100” are not. Note that the garland is not cyclic, i.e. the first turned on lamp is not going after the last turned on lamp and vice versa.

Your task is to find the minimum number of moves you need to make to obtain k-periodic garland from the given one.

You have to answer t independent test cases.

Input
The first line of the input contains one integer t (1≤t≤25 000) — the number of test cases. Then t test cases follow.

The first line of the test case contains two integers n and k (1≤n≤106;1≤k≤n) — the length of s and the required period. The second line of the test case contains the string s consisting of n characters ‘0’ and ‘1’.

It is guaranteed that the sum of n over all test cases does not exceed 106 (∑n≤106).

Output
For each test case, print the answer — the minimum number of moves you need to make to obtain k-periodic garland from the given one.

Example
Input
6
9 2
010001010
9 3
111100000
7 4
1111111
10 3
1001110101
1 1
1
1 1
0
Output
1
2
5
4
0
0
思路：dp不會寫，只能考慮貪心了。
我們將所有的位置都變為0.因為k個位置中，1的數量最多是1個。所以我們枚舉這個位置，然后不斷的更新最小值。具體看代碼：

#include<bits/stdc++.h> #define ll long long #define inf 0x3f3f3f3f using namespace std;string s; int n,k;int main() {int t;scanf("%d",&t);while(t--){scanf("%d%d",&n,&k);cin>>s;int ans=0;int sum=0;for(int i=0;i<n;i++) sum+=(s[i]=='1');ans=inf;ans=min(ans,sum);for(int i=0;i<k;i++){int cnt=0;for(int j=i;j<n;j+=k){if(s[j]=='1') cnt--;else cnt++;cnt=min(cnt,0);//如果cnt>0的話，說明，這個位置上，1的個數小于0的個數，那么我們就沒必要將0換位1了，可以采用將1換為0的策略。這正是貪心的體現。ans=min(ans,sum+cnt);}}cout<<ans<<endl;}return 0; }

努力加油a啊，(^o)/~

總結

以上是生活随笔為你收集整理的K-periodic Garland CodeForces - 1353E（贪心）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。