直接二分答案 然后判断是否符合要求 复杂度n*(log(n*1000))
#include <string.h>
#include <iostream>
#include <stdio.h>
#include <map>
using namespace std;
int ti[10001];
int n,k;
int main(void){
// freopen("","r",stdin);
while(scanf("%d %d",&n,&k)!=EOF){
if(n==0&&k==0) break;
ti[0] = 0;
for(int i = 1;i<=n;i++)
scanf("%d",&ti[i]);
int low = 0,high = n*1000;
while(low<=high){
int mid = (low+high)>>1;
bool ok = true;
int cnt = k;
int sum = 0;
for(int i = 1;i<=n;i++){
if(mid<ti[i]){
ok = false;
break;
}//表示mid不可能完成ti[i]这个任务
if(mid<sum+ti[i]){
sum = 0;
cnt--;
if(cnt<=0){
ok = false;
break;
}//表示mid时间下k个服务器不够
}
sum+=ti[i];
}
if(!ok) low = mid+1;
else high = mid-1;
}
printf("%d\n",low);
}
}