Gangster

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on 

HDU. Original ID: 

64-bit integer IO format: %I64d      Java class name: Main

There are two groups of gangsters fighting with each other. The first group stands in a line, but the other group has a magic gun that can shoot a range [a, b], and everyone in that range will take a damage of c points. When a gangster is taking damage, if he has already taken at least P point of damage, then the damage will be doubled. You are required to calculate the damage that each gangster in the first group toke.

To simplify the problem, you are given an array A of length N and a magic number P. Initially, all the elements in this array are 0.

Now, you have to perform a sequence of operation. Each operation is represented as (a, b, c), which means: For each A[i] (a <= i <= b), if A[i] < P, then A[i] will be A[i] + c, else A[i] will be A[i] + c * 2.

Compute all the elements in this array when all the operations finish.

 

Input

The input consists several testcases.

The first line contains three integers n, m, P (1 <= n, m, P <= 200000), denoting the size of the array, the number of operations and the magic number.

Next m lines represent the operations. Each operation consists of three integers a; b and c (1 <= a <= b <= n, 1 <= c <= 20).

 

Output

Print A[1] to A[n] in one line. All the numbers are separated by a space.

 

Sample Input

3 2 11 2 12 3 1

Sample Output

1 3 1

Source

 

解题：线段树。。时间卡死人啊。。。多交几次g++就过了

 

1 #include 
        
          2 #include 
         
           3 #include 
          
            4 #include 
           
             5 using namespace std; 6 const int maxn = 200010; 7 struct node { 8     int minv,maxv,lazy; 9 } tree[maxn<<2];10 int n,m,p;11 inline void pushdown(int v) {12     if(tree[v].lazy) {13         tree[v<<1].lazy += tree[v].lazy;14         tree[v<<1|1].lazy += tree[v].lazy;15         tree[v<<1].minv += tree[v].lazy;16         tree[v<<1|1].minv += tree[v].lazy;17         tree[v<<1].maxv += tree[v].lazy;18         tree[v<<1|1].maxv += tree[v].lazy;19         tree[v].lazy = 0;20     }21 }22 inline void pushup(int v) {23     tree[v].maxv = max(tree[v<<1].maxv,tree[v<<1|1].maxv);24     tree[v].minv = min(tree[v<<1].minv,tree[v<<1|1].minv);25 }26 void update(int L,int R,int lt,int rt,int val,int v) {27     if(lt <= L && rt >= R && (tree[v].minv >= p || tree[v].maxv < p)) {28         val <<= (tree[v].minv >= p);29         tree[v].minv += val;30         tree[v].maxv += val;31         tree[v].lazy += val;32         return;33     }34     pushdown(v);35     int mid = (L + R)>>1;36     if(lt <= mid) update(L,mid,lt,rt,val,v<<1);37     if(rt > mid) update(mid+1,R,lt,rt,val,v<<1|1);38     pushup(v);39 }40 void query(int L,int R,int v){41     if(L == R){42         if(L > 1) putchar(' ');43         printf("%d",tree[v].lazy);44         return;45     }46     pushdown(v);47     int mid = (L + R)>>1;48     query(L,mid,v<<1);49     query(mid+1,R,v<<1|1);50 }51 int main() {52     int a,b,c;53     while(~scanf("%d %d %d",&n,&m,&p)){54         memset(tree,0,sizeof tree);55         while(m--){56             scanf("%d %d %d",&a,&b,&c);57             update(1,n,a,b,c,1);58         }59         query(1,n,1);60         putchar('\n');61     }62     return 0;63 }