Inna loves sweets very much. She has n closed present boxes lines up in a row in front of her. Each of these boxes contains either a candy (Dima's work) or nothing (Sereja's work). Let's assume that the boxes are numbered from 1 to n, from left to right.
As the boxes are closed, Inna doesn't know which boxes contain candies and which boxes contain nothing. Inna chose number k and asked w questions to Dima to find that out. Each question is characterised by two integers li, ri (1 ≤ li ≤ ri ≤ n; r - l + 1 is divisible byk), the i-th question is: "Dima, is that true that among the boxes with numbers from li to ri, inclusive, the candies lie only in boxes with numbers li + k - 1, li + 2k - 1, li + 3k - 1, ..., ri?"
Dima hates to say "no" to Inna. That's why he wonders, what number of actions he will have to make for each question to make the answer to the question positive. In one action, Dima can either secretly take the candy from any box or put a candy to any box (Dima has infinitely many candies). Help Dima count the number of actions for each Inna's question.
Please note that Dima doesn't change the array during Inna's questions. That's why when you calculate the number of operations for the current question, please assume that the sequence of boxes didn't change.
The first line of the input contains three integers n, k and w (1 ≤ k ≤ min(n, 10), 1 ≤ n, w ≤ 105). The second line contains ncharacters. If the i-th box contains a candy, the i-th character of the line equals 1, otherwise it equals 0.
Each of the following w lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n) — the description of the i-th question. It is guaranteed that ri - li + 1 is divisible by k.
For each question, print a single number on a single line — the minimum number of operations Dima needs to make the answer to the question positive.
10 3 3
1010100011
1 3
1 6
4 9
1
3
2
For the first question, you need to take a candy from the first box to make the answer positive. So the answer is 1.
For the second question, you need to take a candy from the first box, take a candy from the fifth box and put a candy to the sixth box. The answer is 3.
For the third question, you need to take a candy from the fifth box and put it to the sixth box. The answer is 2.
题意：
给你n个礼物盒，一个k，和m次询问
每个礼物盒里边可能放了礼物或者没有礼物(0/1);
也就是一个01串
每次询问你 l,r
其中 l+k-1,l+2*k-1,l+3*k-1........r 有多少个是0 ，另外的位置 有多少个是1
题解：
k最多是10，建立最多10个BIT
我们把每个位置的情况放入 i%k 这个树状数组中
查询的时候查找 (l-1)%k 这个BIT就行了
l,r礼物总和可以用前缀解决
#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18;
const double Pi = acos(-1.0);
const int N = 1e5+, M = 1e6, mod = 1e9+, inf = 2e9; int n,k,m,sum[N];
char a[N];
int C[][N];
void update(int f,int x,int c) {
for(int i = x; i < N; i += i&(-i)) C[f][i] += c;
}
int ask(int f,int x) {
int s = ;
for(int i = x; i; i -= i&(-i)) s += C[f][i];
return s;
}
int main() {
scanf("%d%d%d",&n,&k,&m);
scanf("%s",a+);
for(int i = ; i <= n; ++i) update((i)%k,i,a[i]-''),sum[i] = sum[i-] + a[i] - '';
for(int i = ; i <= m; ++i) {
int l,r;
scanf("%d%d",&l,&r);
int ss = (ask((l-)%k,r) - ask((l-)%k,l-));
printf("%d\n",sum[r] - sum[l-] - ss + (r-l+)/k - ss);
}
return ;
}