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

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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 nand 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 31 2 3 4 51 2 3 4 5 61 1 101 1 -91 5 -1

output

0900

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：求最小的f。

f[i]定义已给出，只是条件多给了一个查询，给l,r区间添加一个x.

问每次添加了x之后的最小f.

显然对于a[i]  b[i]  b[i]的奇加偶减或者  奇减偶加是不会变的，所以我可以预处理出一个sum[i]  （sum[i]的含义看代码）

对于a[i]  我只需先奇加偶减处理出ans=sigma (i&1)?1:-1 a[i]，因为对于a[i]数组来说，他是始终要从1-n加到尾的。

所以我遍历1-n a[i] 一次就好了。

每次查询最小值二分就好了。

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <stdio.h>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>#include <stdlib.h>#include <time.h>#include <bitset>using namespace std;#define pi acos(-1)#define s_1(x) scanf("%d",&x)#define s_2(x,y) scanf("%d%d",&x,&y)#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define PI acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define fOR(n,x,i) for(int i=n;i>=x;i--)#define fOr(n,x,i) for(int i=n;i>x;i--)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");#define db double#define ll long long#define mp make_pair#define pb push_backtypedef pair<long long int,long long int> ii;typedef long long LL;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,1,-1,-1,1};const int dy[]={0,1,0,-1,-1,1,-1,1};const int maxn=1e5+10;const int maxx=600005;const double EPS=1e-8;const double eps=1e-8;const int mod=1e9+7;template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}template <class T>inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}if(IsN) num=-num;return true;}void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}void print(LL a){ Out(a),puts("");}//freopen( "in.txt" , "r" , stdin );//freopen( "data.txt" , "w" , stdout );//cerr << "run time is " << clock() << endl;LL sum[maxn],sum1[maxn],sum2[maxn];LL a[maxn],b[maxn];LL ans=0;int tot;int n,m,q;LL solve(){    int l=1,r=tot;    int len=100;    LL val=LINF;    W(l<r)    {        int mid=l+r+1>>1;        if(ans-sum[mid]>=0) l=mid;        else r=mid-1;    }    //cout<<l<<endl;    val=min(val,abs(ans-sum[l]));    if(l+1<=tot)        val=min(val,abs(ans-sum[l+1]));    return val;}int main(){    W(s_3(n,m,q)!=EOF)    {        ans=0;        LL cur=1;        FOR(1,n,i)        {            scan_d(a[i]);            ans+=cur*a[i];            cur*=-1;        }        sum1[0]=0;        sum2[0]=0;        FOR(1,m,i)        {            scan_d(b[i]);            if(i&1)            {                if(i==1)                {                    sum1[i]=b[i];                }                else sum1[i]=sum1[i-2]+b[i];                sum2[i]=sum2[i-1];            }            else            {                sum2[i]=sum2[i-2]+b[i];                sum1[i]=sum1[i-1];            }        }        FOR(1,m-n+1,i)        {            int r=i+n-1;            if(i&1)            {                sum[i]=sum1[r]-sum1[i-1]-(sum2[r]-sum2[i-1]);            }            else            {                sum[i]=sum2[r]-sum2[i-1]-(sum1[r]-sum1[i-1]);            }        }        tot=m-n+1;        sort(sum+1,sum+1+tot);        print(solve());        W(q--)        {            int l,r;            LL x;            s_2(l,r);            scan_d(x);            if((r-l+1)&1)            {                if(l&1) ans+=x;                else ans-=x;            }            print(solve());        }    }}