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

一開始有很多答案

f0,f1,f2,f3,f4…,fm - n這個我們預處理出來這裡是沒有絕對值的

每次修改l,r 其實隻修改了a的總和

假設修改了總值v

那麼要求的最小的原式 = abs(v+fj)

記錄一下改變的總和然後二分一下找最接近的fj

#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <queue>
#include <cstdio>
#include <map>
#include <set>
#include <utility>
#include <stack>
#include <cstring>
#include <cmath>
#include <vector>
#include <ctime>
#include <bitset>
using namespace std;
#define pb push_back
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define ss(str) scanf("%s",str)
#define ans() printf("%d",ans)
#define ansn() printf("%d\n",ans)
#define anss() printf("%d ",ans)
#define lans() printf("%lld",ans)
#define lanss() printf("%lld ",ans)
#define lansn() printf("%lld\n",ans)
#define fansn() printf("%.10f\n",ans)
#define r0(i,n) for(int i=0;i<(n);++i)
#define r1(i,e) for(int i=1;i<=e;++i)
#define rn(i,e) for(int i=e;i>=1;--i)
#define rsz(i,v) for(int i=0;i<(int)v.size();++i)
#define szz(x) ((int)x.size())
#define mst(abc,bca) memset(abc,bca,sizeof abc)
#define lowbit(a) (a&(-a))
#define all(a) a.begin(),a.end()
#define pii pair<int,int>
#define pli pair<ll,int>
#define mp make_pair
#define lrt rt<<1
#define rrt rt<<1|1
#define X first
#define Y second
#define PI (acos(-1.0))
#define sqr(a) ((a)*(a))
typedef long long ll;
typedef unsigned long long ull;
const ll mod = +;
const double eps=;
const int inf=;
const ll infl = ;
const int maxn=  +;
const int maxm = +;
//Pretests passed
int in(int &ret)
{
    char c;
    int sgn ;
    if(c=getchar(),c==EOF)return -;
    while(c!='-'&&(c<'0'||c>'9'))c=getchar();
    sgn = (c=='-')?-:;
    ret = (c=='-')?:(c-'0');
    while(c=getchar(),c>='0'&&c<='9')ret = ret*+(c-'0');
    ret *=sgn;
    return ;
}

int a[maxn];
int b[maxn];
int main()
{
#ifdef LOCAL
    freopen("input.txt","r",stdin);
//    freopen("output.txt","w",stdout);
#endif // LOCAL

    int n,m,q
;    sddd(n,m,q);
    r1(i,n)in(a[i]);
    r1(i,m)in(b[i]);
    ll x = , y = ;
    int op = ;
    r1(i,n)x+=op*a[i],op=-op;
    vector<ll>v;
    op = -;
    int idx = n&?:;
    ll del = ;
    int op2 = -;
    for(int i = ;i<=m;++i)
    {
        if(i>n)
        {
            del+=op2*b[i-n];
            op2 = -op2;
        }
        y+=op*b[i],op = -op;
        if(i>=n)
        {
            ll tmp = y - del;
            if((i-n+)%==)tmp = -tmp;
            v.pb(tmp+x);
        }
    }
    ll tot = ;
    sort(all(v));
    int sz = v.size();
    r0(i,q+)
    {
        ll l=  ,r=,z=;
        if(i)slddd(l,r,z);
        if((r-l+)&)
        {
            if(l&)tot+=z;
            else tot-=z;
        }
        int pos = lower_bound(all(v),-tot)-v.begin();
        if(pos==sz)pos = sz-;
        ll ans = abs(v[pos]+tot);
        if(pos)ans = min(ans,abs(v[pos-]+tot));
        lansn();
    }
    return ;
}