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>0){
     mid=(lb+ub)/2;
     if(C(mid)<=m) ub=mid;
     else lb=mid+1; //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>1){
     mid=(lb+ub)/2;
     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[4]={1,0,-1,0},dy[4]={0,1,0,-1};
const int INF=0x3f3f3f3f;
const ll mod=1e9+7;
ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
const bool AC=true;

int n,m,maxn,sum;
int a[100005];
int C(int x){ //计算额度为x时的最小month数
    int last,cur,total,ans;
    last=cur=0;ans=0;
    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=0;
    rep(i,0,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>0){ //上下限相等时跳出循环
     mid=(lb+ub)/2;
     if(C(mid)<=m) ub=mid;
     else lb=mid+1; //下限小了
     }
    out(ub);
    return 0;
}