Codeforces Round #435 (Div. 2) E. Mahmoud and Ehab and the function[二分]

题目：http://codeforces.com/problemset/problem/862/E

E. Mahmoud and Ehab and the function
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Dr. Evil is interested in math and functions, so he gave Mahmoud and Ehab array a of length n and array b of length m. He introduced a function f(j) which is defined for integers j, which satisfy 0 ≤ j ≤ m - n. Suppose, ci = ai - bi + j. Then f(j) = |c1 - c2 + c3 - c4... cn|. More formally, .

Dr. Evil wants Mahmoud and Ehab to calculate the minimum value of this function over all valid j. They found it a bit easy, so Dr. Evil made their task harder. He will give them q update queries. During each update they should add an integer xi to all elements in a in range [li;ri]i.e. they should add xi to ali, ali + 1, ... , ari and then they should calculate the minimum value of f(j) for all valid j.

Please help Mahmoud and Ehab.

Input

The first line contains three integers n, m and q (1 ≤ n ≤ m ≤ 105, 1 ≤ q ≤ 105) — number of elements in a, number of elements in b and number of queries, respectively.

The second line contains n integers a1, a2, ..., an. ( - 109 ≤ ai ≤ 109) — elements of a.

The third line contains m integers b1, b2, ..., bm. ( - 109 ≤ bi ≤ 109) — elements of b.

Then q lines follow describing the queries. Each of them contains three integers li ri xi (1 ≤ li ≤ ri ≤ n,  - 109 ≤ x ≤ 109) — range to be updated and added value.

Output

The first line should contain the minimum value of the function f before any update.

Then output q lines, the i-th of them should contain the minimum value of the function f after performing the i-th update .

Example
input
5 6 3
1 2 3 4 5
1 2 3 4 5 6
1 1 10
1 1 -9
1 5 -1
output
0
9
0
0
Note

For the first example before any updates it's optimal to choose j = 0, f(0) = |(1 - 1) - (2 - 2) + (3 - 3) - (4 - 4) + (5 - 5)| = |0| = 0.

After the first update a becomes {11, 2, 3, 4, 5} and it's optimal to choose j = 1, f(1) = |(11 - 2) - (2 - 3) + (3 - 4) - (4 - 5) + (5 - 6) = |9| = 9.

After the second update a becomes {2, 2, 3, 4, 5} and it's optimal to choose j = 1, f(1) = |(2 - 2) - (2 - 3) + (3 - 4) - (4 - 5) + (5 - 6)| = |0| = 0.

After the third update a becomes {1, 1, 2, 3, 4} and it's optimal to choose j = 0, f(0) = |(1 - 1) - (1 - 2) + (2 - 3) - (3 - 4) + (4 - 5)| = |0| = 0.

题意：a数组含有n个元素，b数组含有m个元素。含有q次区间修改，每次由公式计算f(j)最小值。

题解：也许是见到最水的一个E...分开计算a和b，a组的数值固定，由于加减相互交错，每次只需要加一次更新的值或者不加。所有可能的b组数值存一个set，每次在set二分搜索和a差值最小的b，做差得到绝对值最小值。

  1 #define _CRT_SECURE_NO_DEPRECATE
  2 #pragma comment(linker, "/STACK:102400000,102400000")
  3 #include<iostream>  
  4 #include<cstdio>  
  5 #include<fstream>  
  6 #include<iomanip>
  7 #include<algorithm>  
  8 #include<cmath>  
  9 #include<deque>  
 10 #include<vector>
 11 #include<bitset>
 12 #include<queue>  
 13 #include<string>  
 14 #include<cstring>  
 15 #include<map>  
 16 #include<stack>  
 17 #include<set>
 18 #include<functional>
 19 #define pii pair<int, int>
 20 #define mod 1000000007
 21 #define mp make_pair
 22 #define pi acos(-1)
 23 #define eps 0.00000001
 24 #define mst(a,i) memset(a,i,sizeof(a))
 25 #define all(n) n.begin(),n.end()
 26 #define lson(x) ((x<<1))  
 27 #define rson(x) ((x<<1)|1) 
 28 #define inf 0x3f3f3f3f
 29 typedef long long ll;
 30 typedef unsigned long long ull;
 31 using namespace std;
 32 const int maxn = 1e5 + 5;
 33 set<ll>b;
 34 ll orib[maxn];
 35 
 36 void getbest(ll suma)
 37 {
 38     auto it = b.lower_bound(suma);
 39     ll best;
 40     if (it != b.end())
 41     {
 42         best = abs(suma - *it);
 43         if (it != b.begin())
 44         {
 45             it--;
 46             best = min(best, abs(suma - *it));
 47         }
 48     }
 49     else
 50     {
 51         it--;
 52         best = abs(suma - *it);
 53     }
 54     cout << best << endl;
 55     return;
 56 }
 57 
 58 int main()
 59 {
 60     ios::sync_with_stdio(false);
 61     cin.tie(0); cout.tie(0);
 62     int i, j,  m, n, q;
 63     cin >> n >> m >> q;
 64     ll suma = 0;
 65     ll flag = 1, k;
 66     for (int i = 1; i <= n; ++i)
 67     {
 68         cin >> k;
 69         suma += k*flag;
 70         flag *= -1;
 71     }
 72     for (int i = 1; i <= m; ++i)
 73         cin >> orib[i];
 74     ll sumb = 0;
 75     flag = 1;
 76     for (int i = 1; i <= n; ++i)
 77     {
 78         sumb += orib[i] * flag;
 79         flag *= -1;
 80     }
 81     b.insert(sumb);
 82     if (n % 2 == 0)flag = -1;
 83     else flag = 1;
 84     for (int i = n + 1; i <= m; ++i)
 85     {
 86         sumb *= -1;
 87         sumb += orib[i - n];
 88         sumb += orib[i] * flag;
 89         b.insert(sumb);
 90     }
 91     int ta, tb, tc;
 92     getbest(suma);
 93     for (int i = 1; i <= q; ++i)
 94     {
 95         cin >> ta >> tb >> tc;
 96         if ((tb - ta + 1) % 2 != 0)
 97         {
 98             if (ta & 1)suma += tc;
 99             else suma -= tc;
100         }
101         getbest(suma);
102     }
103     return 0;
104 }

 

posted @ 2017-09-20 20:51  Meternal  阅读(175)  评论(0)  编辑  收藏  举报