【杭电】[4004]The Frog's Games
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因为可以方便比较
mid是否比res大
(通过计算mid下跳过石头所需的次数)
所以可以对答案进行二分查找
#include<stdio.h>#include<algorithm>using namespace std;int a[500200];int L,n,m;bool judge(int x) { int cnt=0,t=0; for(int i=1; i<=n; i++) { if(a[i]-a[i-1]>x) return false; if(a[i]-t>x) { cnt++; t=a[--i]; } } if(cnt+1<=m) return true; else return false;}int main() { while(scanf("%d %d %d",&L,&n,&m)!=EOF) { for(int i=1; i<=n; i++) scanf("%d",&a[i]); a[0]=0,a[++n]=L; sort(a,a+n+1); int l=0,r=L,res; while(l<=r) { int mid=(l+r)/2; if(judge(mid)) { res=mid; r=mid-1; } else l=mid+1; } printf("%d\n",res); } return 0;}
题目地址:【杭电】[4004]The Frog’s Games
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