簡單的二分

有 n 個石頭，求最短長度的話，二分一下最短長度判斷完成。。。

以下是 AC 程式碼

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
#define ll long long int
const int maxn = 1e6+5;
ll num[maxn];
int t,m,n;
bool judge(int x)
{
    int res = 0, temp = 0;
    for(int i=1;i<=n;i++)
    {
        if(num[i] - temp > x)
            return false;
        while(num[i] - temp <= x && i <= n)
            i++;
        temp = num[--i];
        res ++;
    }
    return res<=m;
}
int main()
{
    while(~scanf("%d%d%d",&t,&n,&m))
    {
        for(int i=1;i<=n;i++)
            scanf("%lld",&num[i]);
        num[++n] = t;
        sort(num+1,num+1+n);
        int l = 0;
        int r = t;
        int mid, ans;
        while(r - l >= 0)
        {
            mid = (r+l) / 2;
            if(judge(mid))
                r = mid - 1,
                ans = mid;
            else
                l = mid + 1;
        }
        printf("%d\n",ans);
    }
    return 0;
}