BZOJ 1012: [JSOI2008]最大数maxnumber 单调队列/线段树

题目链接:

http://www.lydsy.com/JudgeOnline/problem.php?id=1012

题意:

题解:

不用printf输出就RE…

代码:

代码一: 单调队列

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 #define MS(a) memset(a,0,sizeof(a))
 5 #define MP make_pair
 6 #define PB push_back
 7 const int INF = 0x3f3f3f3f;
 8 const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
 9 inline ll read(){
10     ll x=0,f=1;char ch=getchar();
11     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
12     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
13     return x*f;
14 }
15 //////////////////////////////////////////////////////////////////////////
16 const int maxn = 250001;
17 
18 int a[maxn],maxnum[maxn];
19 
20 int main(){
21     int m,d,k=0;
22     int t = 0;
23     scanf("%d%d",&m,&d);
24     for(int i=0; i<m; i++){
25         char op[2]; scanf("%s",op);
26         if(op[0] == 'A'){
27             int x; scanf("%d",&x);
28             a[++t] = (x+k)%d;
29             for(int i=t; i>0; i--){
30                 if(maxnum[i] < a[t]){
31                     maxnum[i] = a[t];
32                 }else{
33                     break;
34                 }
35             }
36         }else{
37             int x; scanf("%d",&x);
38             k = maxnum[t-x+1];
39             printf("%d\n",k);
40         }
41     }
42 
43     return 0;
44 }

 

代码二:线段树

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 #define MS(a) memset(a,0,sizeof(a))
 5 #define MP make_pair
 6 #define PB push_back
 7 const int INF = 0x3f3f3f3f;
 8 const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
 9 inline ll read(){
10     ll x=0,f=1;char ch=getchar();
11     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
12     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
13     return x*f;
14 }
15 //////////////////////////////////////////////////////////////////////////
16 const int maxn = 2e5+10;
17 
18 int n;ll d;
19 struct node{
20     int l,r;
21     ll mx;
22     void update(ll val){
23         mx = max(mx,val);
24     }
25 }tree[maxn*4];
26 
27 void pushup(int rt){
28     tree[rt].mx = max(tree[rt<<1].mx,tree[rt<<1|1].mx);
29 }
30 
31 void build(int rt,int l,int r){
32     tree[rt].l=l,tree[rt].r = r;
33     tree[rt].mx = 0;
34     if(l == r)
35         return ;
36     int mid = (l+r)/2;
37     build(rt<<1,l,mid);
38     build(rt<<1|1,mid+1,r);
39     pushup(rt);
40 }
41 
42 void update(int rt,int pos,ll val){
43     int L = tree[rt].l, R = tree[rt].r;
44     if(L==R && L==pos){
45         tree[rt].mx = val;
46         return ;
47     }
48     int mid = (L+R)/2;
49     if(pos<=mid) update(rt<<1,pos,val);
50     if(pos>mid) update(rt<<1|1,pos,val);
51     pushup(rt);
52 }
53 
54 ll query(int rt,int l,int r){
55     int L = tree[rt].l, R = tree[rt].r;
56     if(l<=L && R<=r)
57         return tree[rt].mx;
58     ll ans = 0;
59     int mid = (L+R)/2;
60     if(l<=mid) ans = max(ans,query(rt<<1,l,r));
61     if(r>mid) ans = max(ans,query(rt<<1|1,l,r));
62     return ans;
63 }
64 
65 int main(){
66     scanf("%d%lld",&n,&d);
67     build(1,1,n);
68     ll ans = 0,v;
69     int cnt = 0;
70     for(int i=0; i<n; i++){
71         char op;
72         scanf(" %c%lld",&op,&v);
73         if(op == 'A'){
74             v = (ans+v)%d;
75             ++cnt;
76             update(1,cnt,v);
77         }else{
78             ans = query(1,cnt-v+1,cnt);
79             printf("%lld\n",ans);
80         }
81     }
82 
83     return 0;
84 }

 

posted @ 2017-03-02 12:31  _yxg123  阅读(115)  评论(0编辑  收藏  举报