HDU5037 贪心

Problem Description

Once upon a time, there is a little frog called Matt. One day, he came to a river.

The river could be considered as an axis.Matt is standing on the left bank now (at position 0). He wants to cross the river, reach the right bank (at position M). But Matt could only jump for at most L units, for example from 0 to L.

As the God of Nature, you must save this poor frog.There are N rocks lying in the river initially. The size of the rock is negligible. So it can be indicated by a point in the axis. Matt can jump to or from a rock as well as the bank.

You don’t want to make the things that easy. So you will put some new rocks into the river such that Matt could jump over the river in maximal steps.And you don’t care the number of rocks you add since you are the God.

Note that Matt is so clever that he always choose the optimal way after you put down all the rocks.

Input

The first line contains only one integer T, which indicates the number of test cases.

For each test case, the first line contains N, M, L (0<=N<=2*10^5,1<=M<=10^9, 1<=L<=10^9).

And in the following N lines, each line contains one integer within (0, M) indicating the position of rock.

Output

For each test case, just output one line “Case #x: y”, where x is the case number (starting from 1) and y is the maximal number of steps Matt should jump.

Sample Input

2
1 10 5
5
2 10 3
3
6

Sample Output

Case #1: 2
Case #2: 4

这破贪心我做了俩小时….
看着感觉是水题掉以轻心….
结果被狠狠地教做人了………..
一开始还没读懂题…
后来才知道他给的是必须走的…
先知道最小周期最大步数是怎么个走法….
因为设定这个人聪明所以他一步肯定想走L…
但是不能让他走l…
所以给他一步放一个，L步放一个…
这样l+1的周期里就有俩了…
然后模一遍看看剩下多少就好了…
要是有剩余就和上一次剩下的加在一起看看够不够L
不够的话就只能下次再说了…
要是够了的话走一步，一开始初始化为L是为了第一次走，那玩意一定剩不下………给他一个站的位置
不然还要特判很麻烦

#include<iostream>#include<queue>#include<algorithm>#include<cstdio>#include<memory.h>#include<string>using namespace std;int tu[1500000];int main(){    int T;    cin >> T;    int u = 0;    while (T--)    {        int n, m, l;        cin >> n >> m >> l;        for (int a = 1;a <= n;a++)scanf("%d", &tu[a]);        tu[++n] = m;        tu[0] = 0;        int xianzai = l;        long long sum = 0;        sort(tu, tu + n);        for (int a = 0;a < n;a++)        {            int cha = tu[a + 1] - tu[a];            int bu = cha /(l+1);            bu *= 2;            int shengyu = cha%(l+1);            if (shengyu + xianzai > l)            {                xianzai = shengyu;                sum++;                sum += bu;            }            else            {                xianzai += shengyu;                sum += bu;            }        }        printf("Case #%d: %lld\n", ++u, sum);    }    return 0;}