Codeforces 954 G. Castle Defense

Posted 2020-10-28 日拱一卒 功不唐捐

http://codeforces.com/problemset/problem/954/G

 

二分答案

检验的时候，从前往后枚举，如果发现某个位置的防御力<二分的值，那么新加的位置肯定是越靠后越好

差分即可

 

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;

#define N 500001

typedef long long LL;

int n,r; 
LL k;

LL sum[N];
LL cf[N];

template<typename T>
void read(T &x)
{
    x=0; char c=getchar();
    while(!isdigit(c)) c=getchar();
    while(isdigit(c)) { x=x*10+c-‘0‘; c=getchar(); }
}

bool check(LL x)
{
    LL rest=k,need;
    LL add=0;
    memset(cf,0,sizeof(cf));
    for(int i=1;i<=n;++i)
    {
        add+=cf[i];
        if(sum[i]+add<x) 
        {
            need=x-sum[i]-add;
            rest-=need;
            if(rest<0) return false;
            cf[i+1]+=need;
            if(i+2*r<n) cf[i+2*r+1]-=need;
        }
    }
    return true;
}

int main()
{
    read(n); read(r); read(k);
    int x;
    for(int i=1;i<=n;++i)
    {
        read(x);
        sum[max(0,i-r)]+=x;
        if(i+r<n) sum[i+r+1]-=x;
    }
    for(int i=1;i<=n;++i) sum[i]+=sum[i-1];
    LL l=0,r=2e18,mid,ans=0;
    while(l<=r)
    {
        mid=l+r>>1;
        if(check(mid)) ans=mid,l=mid+1;
        else r=mid-1;
    }
    cout<<ans;
}