Codeforce Round #228 Div2 C

Codeforces Round #229 (Div. 2)
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Submission	Time	Verdict
5967057	02/13/2014 12:39AM	Accepted
5967053	02/13/2014 12:38AM	Wrong answer on test 11
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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 l_i, r_i (1 ≤ l_i ≤ r_i ≤ n; r - l + 1 is divisible by k), the i-th question is: "Dima, is that true that among the boxes with numbers from l_i to r_i, inclusive, the candies lie only in boxes with numbers l_i + k - 1, l_i + 2k - 1, l_i + 3k - 1, ..., r_i?"

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.

Input

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 n characters. 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 l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the description of the i-th question. It is guaranteed that r_i - l_i + 1 is divisible by k.

Output

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.

Sample test(s)

Input

10 3 3 1010100011 1 3 1 6 4 9

Output

1 3 2

Note

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.

The only programming contests Web 2.0 platform

Server time: 02/13/2014 12:40AM (p1).

 1 #pragma comment(linker,"/STACK:102400000,102400000")
 2 #include <cstdio>
 3 #include <vector>
 4 #include <cmath>
 5 #include <stack>
 6 #include <queue>
 7 #include <cstring>
 8 #include <iostream>
 9 #include <algorithm>
10 using namespace std;
11 #define INF 0x7fffffff
12 #define maxn 200005
13 int n, m, t, y, mi, k ,flag, p, ans;
14 int a[maxn], b[maxn];
15 char s[maxn];
16 int main(){
17     cin >> n >> k >> m >> s;
18     for (int i = 0; i <= n-k; i++){
19         for (int j = 0; j < k-1; j++)
20         if (s[i+j]=='1')a[i]++;
21         if (s[i+k-1]=='0')a[i]++;
22     }
23     for (int i = n-k; i >= 0; i--)b[i]=b[i+k]+a[i];
24     while (m--){
25         int l, r;
26         cin >> l >> r;
27         cout << b[l-1] - b[r] << endl;
28     }
29     return 0;
30 }

View Code

 1 #pragma comment(linker,"/STACK:102400000,102400000")
 2 #include <cstdio>
 3 #include <vector>
 4 #include <cmath>
 5 #include <stack>
 6 #include <queue>
 7 #include <cstring>
 8 #include <iostream>
 9 #include <algorithm>
10 using namespace std;
11 #define INF 0x7fffffff
12 #define maxn 200005
13 int n, m ,t ,b, y, mi, k ,flag, p, ans;
14 int a[maxn], c[maxn], d[11][maxn];
15 char s[maxn];
16 void add1(int x){
17     while (x <= n){
18         c[x]++;
19         x += x&-x;
20     }
21 }
22 int sum1(int x){
23     int res = 0;
24     while (x>0){
25         res += c[x];
26         x -= x&-x;
27     }
28     return res;
29 }
30 void add2(int id,int x){
31     while (x <= n){
32         d[id][x]++;
33         x += x&-x;
34     }
35 }
36 int sum2(int id,int x){
37     int res = 0;
38     while (x>0){
39         res += d[id][x];
40         x -= x&-x;
41     }
42     return res;
43 }
44 int main(){
45     //char ss[] = "0000000000001000000100010010001100100111000001010100000100001";
46     //cout << strlen(ss);
47     cin >> n >> k >> m;
48     scanf("%s", s);
49     for (int i = 1; i <= n; i++){
50         if (s[i-1] == '1'){
51             add1(i);
52         }
53     }
54     for (int i = 0; i < k; i++){
55         for (int j = i; i+j <= n+k; j+=k){
56             if (s[j-1] == '1'){
57                 add2(i, j);
58             }
59         }
60     }
61     while (m--){
62         int l, r;
63         cin >> l >> r;
64         int q = (l - 1) % k;
65         int s1 = sum2(q, r) - sum2(q, l - 1);
66         int s2 = sum1(r) - sum1(l - 1);
67         int s3 = s2 - s1 + (r-l+1)/k - s1;
68         printf("%d\n", s3);
69     }
70     return 0;
71 }

View Code

posted @ 2014-02-13 04:41 HaibaraAi 阅读(123) 评论(0) 收藏举报

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Be careful: there is 50 points penalty for submission which fails the pretests or resubmission (except failure on the first test, denial of judgement or similar verdicts). "Passed pretests" submission verdict doesn't guarantee that the solution is absolutely correct and it will pass system tests.