【二分】hdu 4004

兩次二分，因為題意是求m次能過河的前提下最小跳躍長度，第一次二分是對答案二分，第二次是對石頭二分

那個lower_bound雖好用，不過它只能返回的id的值小於val，也就是說有數組a[0]=0,a[1]=2,a[2]=6,lower_bound(a,a+n,6)-a=1,即a[1]=2,正確是返回2,即a[2]=6，所以這時候應該處理一下。

#include <list>#include <map>#include <set>#include <queue>#include <string>#include <deque>#include <stack>#include <algorithm>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <limits.h>#include <time.h>#include <string.h>using namespace std;#define LL long long#define PI acos(-1.0)#define MAX INT_MAX#define MIN INT_MIN#define eps 1e-8#define FRE freopen("a.txt","r",stdin)#define MOD 1000000007#define N  500005int a[N];int m;int n;int l;bool gao(int x){    int pos=0,id;    int cnt=1;    while(cnt<=m){        pos+=x;        id=lower_bound(a+1,a+2+n,pos)-a-1;        if(pos==a[id+1])id++;        if(a[id]==l){            return true;        }        pos=a[id];        cnt++;    }    return false;}int main(){    while(scanf("%d%d%d",&l,&n,&m)!=EOF){        int i;        for(i=1;i<=n;i++)scanf("%d",&a[i]);        a[i]=l;        sort(a+1,a+n+2);        int ll=0,rr=l,mid;        int ans;        while(ll<=rr){            mid=(ll+rr)>>1;            if(gao(mid)){                ans=mid;                rr=mid-1;            }            else            ll=mid+1;        }        printf("%d\n",ans);    }    return 0;}