POJ 3273 Monthly Expense(二分查找+边界条件)

时间：2021-03-19 02:13:25

POJ 3273 Monthly Expense

　　此题与POJ3258有点类似,一开始把判断条件写错了,wa了两次,二分查找可以有以下两种:

　　

while(ub-lb>){

     mid=(lb+ub)/;

     if(C(mid)<=m) ub=mid;

     else lb=mid+; //此时下限过小

     }

    out(ub);//out(lb)

我一开始是写的下面这种,下面这种要单独判断lb和ub的值,因为用下面这种判断lb,ub都可能成立

while(ub-lb>){

     mid=(lb+ub)/;

     if(C(mid)<=m) ub=mid;

     else lb=mid;

     }

    if(C(lb)<=m){

    out(lb);

    }

    else out(ub);

二分查找的边界判断一定要灵活

/*

* Created:     2016年03月31日 20时32分15秒 星期四

* Author:      Akrusher

*

*/

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <string>

#include <vector>

#include <deque>

#include <list>

#include <set>

#include <map>

#include <stack>

#include <queue>

#include <numeric>

#include <iomanip>

#include <bitset>

#include <sstream>

#include <fstream>

using namespace std;

#define rep(i,a,n) for (int i=a;i<n;i++)

#define per(i,a,n) for (int i=n-1;i>=a;i--)

#define in(n) scanf("%d",&(n))

#define in2(x1,x2) scanf("%d%d",&(x1),&(x2))

#define inll(n) scanf("%I64d",&(n))

#define inll2(x1,x2) scanf("%I64d%I64d",&(x1),&(x2))

#define inlld(n) scanf("%lld",&(n))

#define inlld2(x1,x2) scanf("%lld%lld",&(x1),&(x2))

#define inf(n) scanf("%f",&(n))

#define inf2(x1,x2) scanf("%f%f",&(x1),&(x2))

#define inlf(n) scanf("%lf",&(n))

#define inlf2(x1,x2) scanf("%lf%lf",&(x1),&(x2))

#define inc(str) scanf("%c",&(str))

#define ins(str) scanf("%s",(str))

#define out(x) printf("%d\n",(x))

#define out2(x1,x2) printf("%d %d\n",(x1),(x2))

#define outf(x) printf("%f\n",(x))

#define outlf(x) printf("%lf\n",(x))

#define outlf2(x1,x2) printf("%lf %lf\n",(x1),(x2));

#define outll(x) printf("%I64d\n",(x))

#define outlld(x) printf("%lld\n",(x))

#define outc(str) printf("%c\n",(str))

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define SZ(x) ((int)(x).size())

#define mem(X,Y) memset(X,Y,sizeof(X));

typedef vector<int> vec;

typedef long long ll;

typedef pair<int,int> P;

const int dx[]={,,-,},dy[]={,,,-};

const int INF=0x3f3f3f3f;

const ll mod=1e9+;

ll powmod(ll a,ll b) {ll res=;a%=mod;for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}

const bool AC=true;

int n,m,maxn,sum;

int a[];

int C(int x){ //计算额度为x时的最小month数

    int last,cur,total,ans;

    last=cur=;ans=;

    while(true){  //注意循环终止条件

    if(last>=n) break;

    total=a[last];

    ans++;

    cur++;

    if(cur>=n) break;

    while(total+a[cur]<=x){

    total+=a[cur];

    cur++;

    if(cur>=n) break;

    }

    last=cur;

    }

  return ans;

 }

int main()

{

    in2(n,m);

    maxn=sum=;

    rep(i,,n){

    in(a[i]);

    maxn=max(maxn,a[i]);

    sum+=a[i];

    }

    int lb,ub,mid;

    lb=maxn;//额度下限

     ub=sum;//额度上限

    while(ub-lb>){ //上下限相等时跳出循环

     mid=(lb+ub)/;

     if(C(mid)<=m) ub=mid;

     else lb=mid+; //下限小了

     }

    out(ub);

    return ;

}

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