hdu 6560 思維

The Hermit

Time Limit: 25000/15000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 103 Accepted Submission(s): 69

Problem Description
The Hermit stands alone on the top of a mountain with a lantern in his hand. The snow-capped mountain range symbolises the Hermit’s spiritual achievement, growth and accomplishment. He has chosen this path of self-discovery and, as a result, has reached a heighted state of awareness

dhh loves to listen to radio. There are N radio stations on a number axis, and the i-th station is located at xi = i. The broadcasting scope of the i-th station is radi , which means stations in the interval [i - radi + 1, i + radi - 1] can receive the signal from the i-th station. For some unknown reason, the left boundary that can receive the i-th station’s signal is non-descending, which meansi i - radi + 1 ≤ i + 1 - radi+1 + 1.
Now dhh wants to listen to the radio from station i, and he finds that the station k, satisfying both of the following conditions, can receive perfect signal from the station i:

k < i and station k can receive station i’s signal.

There exists another station j(k ≤ j < i) such that station k and i can both receive the signal from station j and the distance between station k and j is greater than or equal to the distance between station j and i.

Now dhh wonders for each station i, how many stations can receive the perfect signal from station i.

Input
The first line of the input contains one integer T ≤ 20, denoting the number of testcases. Then T testcases follow. For each testcase:
The first line contains one positve integer N.

The second line contains N positive integers rad1, rad2, … , radN .

It’s guaranteed that 1 ≤ N ≤ 106 ,i - radi + 1 ≥ 1 and i + radi - 1 ≤ N

Output
For the k-th testcase, output “Case k: ans” in one line, where ans represents the xor result of answer for each radio station i.
xor is a bitwise operation, which takes two bit patterns of equal length and performs the logical exclusive OR operation on each pair of corresponding bits. The result in each position is 1 if only the first bit is 1 or only the second bit is 1, but will be 0 if both are 0 or both are 1. In this we perform the comparison
of two bits, being 1 if the two bits are different, and 0 if they are the same.

Sample Input
2
7
1 2 3 4 3 2 1
10
1 1 2 3 4 4 3 2 2 1

Sample Output
Case 1: 2
Case 2: 0
Hint

In the first testcase of the example, the number of stations that can receive the perfect signal from each station $i$ is respectively 0, 0, 1, 2, 1, 0, 0 in order, so the answer must be

0 xor 0 xor 1 xor 2 xor 1 xor 0 xor 0 = 2

題意：

在一條直線上有N個基站，座標是從1~N。他們每個基站都有一個信號輻射範圍，是[i-radi+1,i+radi-1]。每個基站信號覆蓋的最左端是非遞減的。對於每個基站都有（不一定有）一些基站是能接受到完美信號的基站。要求滿足下面的條件。
對於基站k來說：

1. k<i
2. 存在另一個j，j滿足k<=j<i,並且dis(k,j)>=dis(j,i)
3. 基站j的信號能夠覆蓋到k和i

思路：

因爲那個最左端非遞減的性質，以及座標是從1~N這兩點。可以推出滿足這些的點的數據只能是：

1. 1 2 3 4 5 6 7 8 遞增不超過1的
2. 2 2 2 2 2 2 2 2 都相同的
3. 10 8 5 3 2 1 這樣遞減的
4. 三者混合起來的

分別對於前三個討論，都能發現除了i基站和i基站前面一個基站不能接受到完美信號，剩下的它能覆蓋到的都能接受到完美信號。

代碼：

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int t,n,x,cas=1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&x);
            if(i>2&&x>2)
                ans^=(x-2);
        }
        printf("Case %d: %d\n",cas++,ans);
    }
    return 0;
}