题目描述
In the shooting game, the player can choose to stand in the position of [1, X] to shoot, you can shoot all the nearest K targets. The value of K may be different on different shootings. There are N targets to shoot, each target occupy the position of [Li, Ri] , the distance from the starting line is Di(1<=i<=N). Shooting a target can get the bonus points equal to the distance to the starting line. After each shooting targets still exist. Additional, if the previous bonus points is more than P, then as a reward for the shooting present bonus points doubled(and then check the next one). Player wants to know the bonus points he got in each shot.
输入
The input consists several test cases.
The first line has two integers, N, M, X, P(1<=N, M ,X<=100000, P<=1000000000), N represents the total number of target shooting, M represents the number of shooting.
The next N lines of three integers L, R, D (1<=L<=R<=X, 1<=D<=10000000), occupy the position of [L, R] and the distance from the starting line is D.
The next M lines, each line contains four integers x, a, b, c (1<=x<=X, 0<=a,b<=N, 1<=c<=10000000), and K = ( a * Pre + b ) % c. Pre is the bonus point of previous shooting , and for the first shooting Pre=1. It denotes standing in the position x and the player can shoot the nearest K targets.
输出
Output M lines each corresponds to one integer.
样例输入
4 3 5 8
1 2 6
2 3 3
2 4 7
1 5 2
2 2 1 5
3 1 1 10
4 2 3 7
样例输出
11 10 18
呼呼呼临走前终于弄出来了，又一道主席树
总之先占个坑先，明天再补全
1 #include<cstdio> 2 #include<algorithm> 3 #define ll long long 4 using namespace std; 5 int N,M,X,P,tot,siz,st; 6 int rt[100005]; 7 struct node{ 8 int l,r,s; 9 ll sum; 10 }t[2000005]; 11 struct mmove{ 12 int x,opt; 13 ll sum; 14 }g[200005]; 15 bool cmp(mmove A,mmove B){return A.x<B.x;} 16 ll ds[100005],pre; 17 int insert(int k,int x,int o,int l,int r){ 18 t[++tot]=t[k];k=tot; 19 if(l==r){ 20 t[k].s+=o; 21 t[k].sum+=ds[l]*o; 22 return k; 23 } 24 int mid=(l+r)/2; 25 if(x<=mid) t[k].l=insert(t[k].l,x,o,l,mid); 26 else t[k].r=insert(t[k].r,x,o,mid+1,r); 27 t[k].s=t[t[k].l].s+t[t[k].r].s; 28 t[k].sum=t[t[k].l].sum+t[t[k].r].sum; 29 return k; 30 } 31 ll query(int k,int x,int l,int r){ 32 if(l==r)return (t[k].s==0?0:(t[k].sum/t[k].s)*min(t[k].s,x)); 33 int mid=(l+r)/2; 34 if(t[t[k].l].s>=x)return query(t[k].l,x,l,mid); 35 else return t[t[k].l].sum+query(t[k].r,x-t[t[k].l].s,mid+1,r); 36 } 37 int main(){ 38 while(scanf("%d%d%d%d",&N,&M,&X,&P)!=EOF){ 39 pre=1;tot=0; 40 t[0]=(node){0,0,0,0}; 41 for(int i=1;i<=N;i++){ 42 scanf("%d%d%lld",&g[i].x,&g[i+N].x,&ds[i]); 43 g[i+N].x++; 44 g[i].opt=1;g[i+N].opt=-1; 45 g[i+N].sum=g[i].sum=ds[i]; 46 } 47 sort(ds+1,ds+1+N); 48 siz=unique(ds+1,ds+1+N)-ds-1; 49 for(int i=1;i<=2*N;i++) 50 g[i].sum=lower_bound(ds+1,ds+1+siz,g[i].sum)-ds; 51 sort(g+1,g+2*N+1,cmp); 52 st=1; 53 for(int i=1;i<=X;i++){ 54 rt[i]=rt[i-1]; 55 while(st<=2*N&&g[st].x==i){ 56 rt[i]=insert(rt[i],g[st].sum,g[st].opt,1,N); 57 st++; 58 } 59 } 60 for(int i=1,x,a,b,c,K;i<=M;i++){ 61 scanf("%d%d%d%d",&x,&a,&b,&c); 62 K=(1ll*a*pre+b)%c; 63 pre=query(rt[x],K,1,N)*(pre>P?2:1); 64 printf("%lld\n",pre); 65 } 66 } 67 }