Monthly Expense POJ

Monthly Expense POJ - 3273 题目地址

题意

Farmer john 知道接下来n天里每天花多少钱, 他想把接下来的n天划分m个fajomonths, 问怎样划分, 使的花费最大的哪一个fajomonths花费最小

分析

1 题目已经说了这么清楚, 肯定是二分搜索啦, 最小化最大值, 二分花费, 只不过本题需要注意的是r = 这n天里花费最大的那一天的花费

参考代码

(好吧,笨笨的我写了好长时间, wrong了好多次)

#include <iostream>#include <cstdio>#define fo0(i,n) for(int i = 0; i < n;+++i)#define fo1(i,n) for(int i = 1; i <= n; ++i)using namespace std;const  int LEN = 100000+4;int cost[LEN];int N,M;bool check(int mid)//判断函数{    long long week = 0;    long long sum = 0;    for(int i = 1;i <= N; ++i)    {        sum+=cost[i];        if(sum>mid)           {               sum = cost[i];               week++;           }       else if(sum == mid)       {           sum = 0;           week++;       }        if(week>M)         return false;    }    if(sum!=0)        ++week;    if(week<=M)    return true;    else    return false;}int main(){    cin>>N>>M;    int Max = 0;    fo1(i,N)   {        scanf("%d",&cost[i]);        if(cost[i]>Max)            Max = cost[i];   }    int l = Max,r = 0x7fffffff;   // cout<<r<<endl;    int mid;    int ans;    while(r-l>1)//二分搜索    {        mid = l+(r-l)/2;        if(check(mid))         {                ans = mid;                r = mid-1;         }        else            l = mid+1;    }    if(check(l))       cout<<l<<endl;    else if(check(r))        cout<<r<<endl;    else        cout<<ans<<endl;    return 0;}