Bzoj 1639: [usaco2007 Mar] monthly expense monthly expenditure

Description

Farmer John is a surprising accounting genius and he understands that he may spend his money to keep the farm running properly every month. He has calculated the cost of moneyi (1 <= moneyi <= 100,000) per day in N (1 <= n <= 10,000) working days ), he wants to make a budget for his consecutive M (1 <= m <= N) checkout period called "liquidation month, each "liquidation month" includes one working day or more consecutive working days, and each working day is only included in one "liquidation month. FJ's goal is to arrange these "liquidation months" to minimize the maximum expense of each liquidation month to determine his monthly expenditure limit. Input

First line: two integers separated by spaces: N and m

2nd. n + 1 row: line I + 1 contains the output of FJ's spending on the I + 1 business day.

Row 1: The amount of money that can maintain normal farm operation every month

Question:

M <= n. If it is divided into <m liquidation months to meet the requirements, it can also be divided into M liquidation months.

The maximum value is the smallest. Two points are considered.

Divide it into two ans.

Then simulate how much money is used up every day,

The expected liquidation month count CNT is obtained.

If CNT <= m is valid, otherwise it is invalid.

Code:

#include<cstdio>#include<cstring>#include<algorithm>//by zrt//problem:using namespace std;typedef long long LL;LL a[100005],n,k;LL maxx=0;LL l,r;bool judge(LL x){    LL sum=0;    LL cnt=0;    for(int i=1;i<=n;i++){        if(sum+a[i]>x){            sum=a[i];            cnt++;            if(cnt>k) return 0;        }else{            sum+=a[i];        }    }    if(sum)cnt++;    if(cnt<=k) return 1;    else return 0;}LL sum;int main(){    #ifdef LOCAL    freopen("in.txt","r",stdin);    freopen("out.txt","w",stdout);    #endif    scanf("%lld%lld",&n,&k);    for(int i=1;i<=n;i++){        scanf("%lld",&a[i]);        maxx=max(maxx,a[i]);        sum+=a[i];    }    l=maxx-1,r=sum;    while(r-l>1){        int m=(l+r)>>1;        if(judge(m)){            r=m;        }else l=m;    }    printf("%lld\n",r);    return 0;}

Bzoj 1639: [usaco2007 Mar] monthly expense monthly expenditure