Traversal
Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 722 Accepted Submission(s):
268
Problem Description
I arrive at a big lake and I have to cross it. Luckily,
I’m a very good jumper, but the lake is too big to be crossed in one jump. On
the shore, I find N boxes of different heights, set in a certain order. If I
throw a box into the lake, it will float and it will have the same height as on
the shore. This is good, because I intend to throw some boxes into the lake and
get from one shore to the other by jumping from box to box. The only things to
consider are:
The lake is big, so I must throw at least 2 boxes, which means that in order to cross the lake I have to make at least 3 jumps.
Not all the boxes have to be thrown; some of them may be ignored.
The boxes can be thrown into the lake only in the order they are found on the shore and I have to jump on them in this order.
The height difference between two consecutive boxes I use must be at most H meters, because I can jump a lot in length, but I have some problems with jumping in height.The height of a box doesn’t change when I jump on it.
I’m always able to jump from the shore to a box and from a box to the shore, no matter what the height of the box is.
Facing so many possibilities that respect the above conditions, I begin counting the number of possibilities that I have, instead of actually crossing the lake. I quickly find the answer and I wonder whether you can also find it as fast as I did.
Task
Write a program that determines the number of possibilities to cross the lake in the above conditions. Since the number can be quite big, you only have to output the remainder of this number, when divided by 9901.
The lake is big, so I must throw at least 2 boxes, which means that in order to cross the lake I have to make at least 3 jumps.
Not all the boxes have to be thrown; some of them may be ignored.
The boxes can be thrown into the lake only in the order they are found on the shore and I have to jump on them in this order.
The height difference between two consecutive boxes I use must be at most H meters, because I can jump a lot in length, but I have some problems with jumping in height.The height of a box doesn’t change when I jump on it.
I’m always able to jump from the shore to a box and from a box to the shore, no matter what the height of the box is.
Facing so many possibilities that respect the above conditions, I begin counting the number of possibilities that I have, instead of actually crossing the lake. I quickly find the answer and I wonder whether you can also find it as fast as I did.
Task
Write a program that determines the number of possibilities to cross the lake in the above conditions. Since the number can be quite big, you only have to output the remainder of this number, when divided by 9901.
Input
There are multiple test cases. Each test case contains
two integers N and H, separated by a space, representing the number of boxes and
the maximum height difference between two consecutive boxes thrown into the
lake. The following N lines contain the heights of the boxes, in the order the
boxes are set on the shore. The (i+1)th line contains the height of the ith box.
Output
For each test case you should output a single line,
containing the number of possibilities modulo 9901.
Constraints
1 < N < 100 001
0 < H < 100 000 001
The height of any box is a strictly positive integer and does not exceed 100 000 000
Constraints
1 < N < 100 001
0 < H < 100 000 001
The height of any box is a strictly positive integer and does not exceed 100 000 000
Sample Input
4 2
1
3
7
5
Sample Output
4
还是太弱，弄一题就得找题解。。。。
1 #include<iostream> 2 #include<string> 3 #include<cstdio> 4 #include<vector> 5 #include<queue> 6 #include<stack> 7 #include<algorithm> 8 #include<cstring> 9 #include<stdlib.h> 10 #include<string> 11 #include<cmath> 12 #include<map> 13 using namespace std; 14 #define mod 9901 15 #define pb push_back 16 #define LL __int64 17 #define mmax 100000+10 18 LL p[mmax],love[mmax],fuck[mmax],cnt,n,m; 19 int search1(LL x){ //找到>=x的第一个数 20 int l=1,r=cnt,mid,tmp=cnt+1; 21 while(l<=r){ 22 mid=(l+r)>>1; 23 if(fuck[mid]>=x) r=mid-1,tmp=min(tmp,mid); 24 else l=mid+1; 25 } 26 return tmp; 27 } 28 int search2(LL x){ //找到<=x的最后一个数 29 int l=1,r=cnt,mid,tmp=0; 30 // cout<<"x="<<x<<endl; 31 while(l<=r){ 32 mid=(l+r)>>1; 33 if(fuck[mid]<=x){ 34 // cout<<"tmp="<<tmp<<endl; 35 l=mid+1,tmp=max(tmp,mid); 36 // cout<<"tmp="<<tmp<<endl; 37 } 38 else r=mid-1; 39 } 40 return tmp; 41 } 42 void update(int pos,int x){ 43 while(pos<=cnt){ 44 p[pos]+=x; 45 p[pos]%=mod; 46 pos+=pos&(-pos); 47 } 48 } 49 int getnum(int pos){ 50 LL sum=0; 51 while(pos>=1){ 52 sum+=p[pos]; 53 pos-=pos&(-pos); 54 } 55 return sum%mod; 56 } 57 int main(){ 58 #ifndef ONLINE_JUDGE 59 freopen("input.txt","r" ,stdin); 60 #endif // ONLINE_JUDGE 61 while(cin>>n>>m){ 62 cnt=0; 63 for(int i=1;i<=n;i++) scanf("%I64d",&love[i]); 64 fuck[++cnt]=love[1]; 65 for(int i=2;i<=n;i++) if(love[i]!=love[i-1]) fuck[++cnt]=love[i]; 66 sort(fuck+1,fuck+1+cnt); 67 LL sum=0;memset(p,0,sizeof(p)); 68 for(int i=1;i<=n;i++){ 69 int a=search1(love[i]-m),b=search2(love[i]+m),c=search1(love[i]); 70 int tmp=getnum(b)-getnum(a-1); 71 //tmp%=mod; 不着为啥下面的改成这样就错了 72 tmp=(tmp%mod+mod)%mod; 73 sum+=tmp; 74 sum%=mod; 75 update(c,tmp+1); 76 } 77 cout<<sum<<endl; 78 } 79 }