P1404 平均数

首先二分答案求出一个可能的最大平均值，并用前缀和数组判断是否有连续至少m个数的平均数超过该可能平均值，之后按二分常规操作就行了

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
#include<deque>
#define ll long long
using namespace std;
const int oo=0x3f3f3f3f;
const int N=100005;

int n,m;
ll r,l,mid,M;
ll a[N],f[N];

ll Max(ll a,ll b){return a>b?a:b;}
ll Min(ll a,ll b){return a<b?a:b;}
ll Abs(ll a){return a>0?a:-a;}

ll get(){
    ll zy=getchar();
    ll z=1,y=0;
    while(zy>'9'||zy<'0'){
        if(zy=='-') z=-1;
        zy=getchar();
    }
    while(zy>='0'&&zy<='9'){
        y=(y<<1)+(y<<3)+zy-'0';
        zy=getchar();
    }
    return z*y;
}

bool likezy(){
    bool ok=false;
    ll p=M;
    for(int i=1;i<=n;i++){
        f[i]=f[i-1]+a[i]-mid;
        if(i>=m){
            p=Min(p,f[i-m]);
            if(f[i]>=p) return true;
        }
    }
    return false;
}

int main(){
    //freopen("P1404平均数.in","r",stdin);
    //freopen("P1404平均数.out","w",stdout);
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++){
        a[i]=get()*10000;
        r=Max(a[i],r);
        M=r;
    }
    while(l<r){
        mid=(l+r)>>1;
        if(likezy()) l=mid+1;
        else r=mid-1;
    }
    printf("%lld
",l/10);
    return 0;
}