題目描述
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.
Hehas chosen this path of self-discovery and， asa result, has reached a heighted state of awareness.

dhh loves to listen to radio. There are 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 means 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 and 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.

輸入
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

輸出
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.

樣例輸入
2
7
1 2 3 4 3 2 1
10
1 1 2 3 4 4 3 2 2 1

樣例輸出
Case 1: 2
Case 2: 0

提示
In the first testcase of t he example, t he number of stations that can receive t he 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

思路
因為每個基站的輻射的左邊界是非下降的，因此每個基站能接受到其信號的基站數是該基站的輻射范圍減去兩邊邊界與該基站位置減去兩邊邊界的最小值

代碼實現

#include<cstdio> #include<algorithm> using namespace std;int main() {int T;scanf("%d",&T);for(int id=1;id<=T;id++){int n;int ans=0;scanf("%d",&n);for(int i=1;i<=n;i++){int temp;scanf("%d",&temp);temp=max(0,min(i-2,temp-2));ans^=temp;}printf("Case %d: %d\n",id,ans);}return 0; }

總結

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