poj 3273 （二分法逼近）

Farmer John is an astounding accounting wizard and has realized he might run out of money to run the farm. He has already calculated and recorded the exact amount of money (1 ≤ money_i ≤ 10,000) that he will need to spend each day over the next N (1 ≤ N ≤ 100,000) days.

FJ wants to create a budget for a sequential set of exactly M (1 ≤ M ≤ N) fiscal periods called "fajomonths". Each of these fajomonths contains a set of 1 or more consecutive days. Every day is contained in exactly one fajomonth.

FJ's goal is to arrange the fajomonths so as to minimize the expenses of the fajomonth with the highest spending and thus determine his monthly spending limit.

Input

Line 1: Two space-separated integers: N and M 
Lines 2.. N+1: Line i+1 contains the number of dollars Farmer John spends on the ith day

Output

Line 1: The smallest possible monthly limit Farmer John can afford to live with.

Sample Input

7 5100400300100500101400

Sample Output

500

枚举区间最大值，然后用二分法逼近

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <stack>#include <map>#include <set>#include <vector>#include <queue>#define mem(p,k) memset(p,k,sizeof(p));#define rep(i,j,k) for(int i=j; i<k; i++)#define pb push_back#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define inf 0x6fffffff#define ll long longusing namespace std;const int mod=1e9+7;char s[2100];int bk[220],f[50000];int n,m,k,minn,maxx,ans,cur;ll x[201000],y[2100];int nex[4][2]={0,1,1,0,0,-1,-1,0};int nextt[101000];int ss(ll mm){    int le=1;    ll sum=0;    rep(i,0,n){        sum+=x[i];//cout<<sum<<endl;        if(sum>mm){            sum=x[i];            le++;        }    }    return le;}ll kmax(ll a,ll b){    return a>b?a:b;}int main(){    while(cin>>n>>m){        maxx=0;        rep(i,0,n)scanf("%lld",x+i),maxx=kmax(maxx,x[i]);        ll l=maxx,r=1e9+2;        while(l<=r){            ll mm=(l+r)>>1;            //cout<<l<<' '<<r<<endl;            int k=ss(mm);            //cout<<l<<"  "<<r<<endl;            if(k<=m){                r=mm-1;            }            else l=mm+1;        }        cout<<l<<endl;    }    return 0;}//freopen("C:\\Users\\LENOVO\\Desktop\\read.txt","r",stdin);