链接:https://www.nowcoder.com/acm/contest/106/K
来源:牛客网
题目描述
It’s universally acknowledged that there’re innumerable trees in the campus of HUST.
Now you're going to walk through a large forest. There is a path consisting of N stones winding its way to the other side of the forest. Between every two stones there is a distance. Let d
i indicates the distance between the stone i and i+1.Initially you stand at the first stone, and your target is the N-th stone. You must stand in a stone all the time, and you can stride over arbitrary number of stones in one step. If you stepped from the stone i to the stone j, you stride a span of (d
i+d
i+1+...+d
j-1). But there is a limitation. You're so tired that you want to walk through the forest in no more than K steps. And to walk more comfortably, you have to minimize the distance of largest step.
输入描述:
The first line contains two integer N and K
as described above.
Then the next line N-1 positive integer followed, indicating the distance between two adjacent stone
.
输出描述:
An integer, the minimum distance of the largest step.
示例1
输入
6 3 1 3 2 2 5
输出
5
题解:
二分答案(好蠢啊)
/* data:2018.5.22 author:gsw link:https://www.nowcoder.com/acm/contest/106/K */ #define ll long long #define IO ios::sync_with_stdio(false); #include<math.h> #include<stdio.h> #include<string.h> #include<iostream> #include<algorithm> using namespace std; #define maxn 100005 ll l,r,mid; int n,k; ll dis[maxn]; bool judge(ll d) { ll tem=0;int kk=0; for(int i=0;i<n-1;i++) { if(dis[i]>d)return false; if(tem+dis[i]>d) { tem=dis[i]; kk++; } else tem+=dis[i]; } kk++; if(kk>k)return false; return true; } int main() { l=r=0; scanf("%d%d",&n,&k); for(int i=0;i<n-1;i++) { scanf("%lld",&dis[i]); r+=dis[i]; }ll mid=0; while(l<r-1) { mid=(l+r)>>1; if(judge(mid))r=mid; else l=mid; } printf("%lld\n",r); return 0; }