Gym - 101028I March Rain 二分

http://codeforces.com/gym/101028/problem/I

题意：给你n个洞，k个布条，问布条能贴到所有洞时的最小值。

题解：二分答案，因为答案越大就越容易满足条件。

技巧：两种judge写法：常规与upper_bound，嗯，就是有种可以upper_bound的感觉。

坑：VS提示upper_bound出错但是代码能ac Orz//_DEBUG_ERROR2("invalid iterator range", _File, _Line);百度不到为啥。。最后发现

 pos = upper_bound(a + pos + 1, a + n + 1, a[pos] + mid - 1) - a;改成

 pos = upper_bound(a + pos , a + n + 1, a[pos] + mid - 1) - a;就好了。有点意思

ac代码：正常写法

#define    _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<vector>
#include<algorithm>
#include<stdio.h>
using namespace std;
const int maxn = 1e5 + ;
int a[maxn];
int n, k;
bool judge(int x) {
    int p = , cnt = ;
    for (int i = ; i <= n; i++) {
        if (p < a[i]) {
            cnt++;
            p = a[i] + x - ;
            if (cnt == k) {
                if (p >= a[n])return ;
                else return ;
            }
            if (p >= a[n])return ;
        }
    }
    return ;
}
int main() {
    int t;
    cin >> t;
    while (t--) {

        cin >> n >> k;
        for (int i = ; i <= n; i++) {
            scanf("%d", &a[i]);
        }
        int l = , r = a[n];
        int mid;
        while (l <= r) {
            mid = l + r >> ;
            if (!judge(mid)) l = mid+;
            else r = mid-;

        }
        cout << l << endl;
    }

}

简化版：

#define    _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<vector>
#include<algorithm>
#include<stdio.h>
using namespace std;
const int maxn = 1e5 + ;
int a[maxn];
int n, k;
bool judge(int x) {
    int pos = , cnt = ;
    for (int i = ; i <= k; i++) {
         pos = upper_bound(a + pos + , a + n + , a[pos] + x - ) - a;
    }
    if (pos > n)return ;
    return ;
}
int main() {
    int t;
    cin >> t;
    while (t--) {

        cin >> n >> k;
        for (int i = ; i <= n; i++) {
            scanf("%d", &a[i]);
        }
        int l = , r = a[n];
        int mid;
        while (l <= r) {
            mid = l + r >> ;
            int pos = ;
            for (int i = ; i <= k; i++) {
                pos = upper_bound(a + pos + , a + n + , a[pos] + mid - ) - a;
            }
            if (pos>n) r = mid - ;
            else l = mid + ;

        }
        cout << l << endl;
    }

}