poj 3273

剛開始沒有注意到 每個月都得在一組裏面，即當a[i]>當前的最大值時，是不符合條件的。
答案的話明顯用二分，二分那個最大花費就行。

#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<time.h>
#include<queue>
#include<stack>
#include<iterator>
#include<math.h>
#include<stdlib.h>
#include<limits.h>
#include<map>
#include<set>
#include<bitset>
//#define ONLINE_JUDGE
#define eps 1e-5
#define INF 0x7fffffff
#define FOR(i,a) for((i)=0;i<(a);(i)++)
#define MEM(a) (memset((a),0,sizeof(a)))
#define sfs(a) scanf("%s",a)
#define sf(a) scanf("%d",&a)
#define sfI(a) scanf("%I64d",&a)
#define pf(a) printf("%d\n",a)
#define pfI(a) printf("%I64d\n",a)
#define pfs(a) printf("%s\n",a)
#define sfd(a,b) scanf("%d%d",&a,&b)
#define sft(a,b,num) scanf("%d%d%d",&a,&b,&num)
#define for1(i,a,b) for(int i=(a);i<b;i++)
#define for2(i,a,b) for(int i=(a);i<=b;i++)
#define for3(i,a,b)for(int i=(b);i>=a;i--)
#define MEM1(a) memset(a,0,sizeof(a))
#define MEM2(a) memset(a,-1,sizeof(a))
#define ll long long
const double PI=acos(-1.0);
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
using namespace std;
//#pragma comment(linker,"/STACK:1024000000,1024000000")
int n,m,L;
#define N 210
#define M 1000100
#define Mod 1000000000
#define p(x,y) make_pair(x,y)
const int MAX_len=550;
int f[M];
int a[100100];
bool check(int x){
    int res = 0;
    int group=1;
    for(int i=0;i<n;i++){
        if(a[i]>x) return false;    //注意這個條件，每個都得在一組裏面
        if(res+a[i]<=x){        //和<=x的爲一組
            res += a[i];
        }else{
            res = a[i];
            group++;
        }
    }
    return (group<=m);
}
int main(){
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
//  freopen("out.txt", "w", stdout);
#endif
    while(sfd(n,m)!=EOF){
        for(int i=0;i<n;i++){
            sf(a[i]);
        }
        int l=0,r=INF;
        while(l<=r){
            int mid = (l+r)>>1;
            if(check(mid))
                r = mid-1;
            else
                l = mid+1;
        }
        printf("%d\n",l);
    }
    return 0;
}