poj 3273

来源：互联网发布：软件质量管理制度编辑：程序博客网时间：2024/06/08 08:20

刚开始没有注意到每个月都得在一组里面，即当a[i]>当前的最大值时，是不符合条件的。
答案的话明显用二分，二分那个最大花费就行。

#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 longconst 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;    //注意这个条件，每个都得在一组里面        if(res+a[i]<=x){        //和<=x的为一组            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;}

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