Codeforces862 E. Mahmoud and Ehab and the function

传送门

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sumA=∑ni=1(−1)i−1ai
sumB[j]=∑ni=1(−1)ibi+j
显然f(j)=|sumA+sumB[j]|

对每次更新[l,r,x]
显然只有(r−l+1)%2==1时,sumA才会改变
当l%2==1,sumA+=x
否则sumA−=x

而sumB[j]=−(sumB[j−1]+bj)+bn+j∗(n%2==1?−1:1)
所以sumB可以在O(m)处理出来
sumA每次update可以O(1)处理
二分找最接近sumA的sumB即可得到答案

#include<stdio.h>#include<bits/stdc++.h>#define ll long long#define pii pair<int,int>#define pll pair<ll,ll>#define MEM(a,x) memset(a,x,sizeof(a))#define lowbit(x) ((x)&-(x))using namespace std;const int INF = 1e9+7;const int inf=INF;const int N = 1e5+50;ll sumA;int b[N];ll sumB[N];void getSumB(int m,int n){    sumB[0]=0;    for(int i=1,flag=-1;i<=m;++i,flag*=-1){        scanf("%d",&b[i]);        if(i<=n){            sumB[0]+=b[i]*flag;        }    }    for(int i=1;n+i<=m;++i){        sumB[i]=-(sumB[i-1]+b[i])+b[n+i]*((n&1)?-1:1);    }}void getSumA(int n){    sumA=0;    for(int i=1,flag=1;i<=n;++i,flag*=-1){        int x;        scanf("%d",&x);        sumA+=flag*x;    }}void add(int l,int r,int x){    if((r-l)%2==0){        if(l%2==1){            sumA+=x;        }        else{            sumA-=x;        }    }}ll __abs(ll x){    return x>0?x:-x;}ll query(int l,int r,int x,int bSz){    add(l,r,x);    ll*it=lower_bound(sumB,sumB+bSz,-sumA);    ll ans=1e18+5;    if(it!=sumB+bSz){        ans=min(ans,__abs(sumA+*it));    }    if(it!=sumB){        --it;        ans=min(ans,__abs(sumA+*it));    }    //add(l,r,-x);    return ans;}int main(){    //freopen("/home/lu/code/r.txt","r",stdin);    //freopen("/home/lu/code/w.txt","w",stdout);    int n,m,q;    while(~scanf("%d%d%d",&n,&m,&q)){        getSumA(n);        getSumB(m,n);        int bSz=m-n+1;        sort(sumB,sumB+bSz);        printf("%lld\n",query(1,1,0,bSz));        while(q--){            int l,r,x;            scanf("%d%d%d",&l,&r,&x);            printf("%lld\n",query(l,r,x,bSz));        }    }    return 0;}