stone二分答案

#include<iostream>#include<fstream>#include<algorithm>#include<stdio.h>using namespace std;ifstream fin("stone.in");ofstream fout("stone.out");#define cin fin#define cout foutint Sstone[50005],Nstone[50005];int N,M,L,l,r;bool pass(int Min){int _use=0;for (int i=1;i<=N;i++) Nstone[i]=Sstone[i];  //将原来的一组距离复制一遍for (int i=1;i<=N;i++)if (Nstone[i]<Min)  //如果距离小于二分的最小值，就移走一块石头。{if (_use>=M) return false;//如果经费不够，方案就不通过else{_use++;Nstone[i+1]+=Nstone[i]; //与后面的合并}}return true;}int main(){int now,last=0;cin >> L >> N >> M;r=L;for (int i=1;i<=N;i++){cin >> now;Sstone[i]=now-last;last=now;}N++;Sstone[N]=L-last;while (l<=r){if (pass((l+r)/2)) l=(l+r)/2+1;  //二分，如果成立，放入左值，否则放入右值else r=(l+r)/2-1;}cout << l-1;}