P1182 数列分段 Section II

题目分析：

  1. 数据大小
  2. 把长度为N的数列分为M段，找出每段和的最大值中的最小值
  3. 二分法：找出答案区间每段和最大值[l,r]
  4. 答案存在一个分界点，[l,x] 为非可取解，x为最优解,[x + 1,r]为可取解
  5. 区间左端点为数列的最大值，右端点为数列的和
  6. 整数二分解题,mid取最小情况(l + r) >> 1;

代码如下:

#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<set>

using namespace std;

#define inf 0x3f3f3f3f
#define eps 1e-6
#define db double
#define ll long long
#define  mm(a,x) memset(a,x,sizeof a)
#define pii pair<int,int>
#define pb push_back
#define el endl
#define debug(x) cerr<<#x<<" = "<<x<<endl
#define fgx cerr<<"-------------------------"<<endl
#define shutio ios::sync_with_stdio(false),cin.tie(0)
#define  mk make_pair
#define lowbit(x) (x) & (-x)
#define fi first
#define se second

const int N = 1e5 + 10;

ll n,m;
ll a[N];
ll l,r;

bool check(int x){
    ll tot = 1;
    ll s = 0;
    for(int i = 0; i < n; i ++ ){
        if(s + a[i] > x){
            s = 0;
            tot ++; 
        }
        s += a[i];
    }
    return tot <= m;    
}

int main() {
    shutio;
    cin >> n >> m;
    for(int i = 0; i < n; i ++ ){
        cin >> a[i];
        l = max(l,a[i]);
        r += a[i];        
    }
    while(l < r){
        int mid = (l + r) >> 1;
        if(check(mid)) r = mid;
        else l = mid + 1;
    }
    cout<<l;
    return 0;
}