To the moon
Problem Description
Background To The Moon is a independent game released in November 2011, it is a role-playing adventure game powered by RPG Maker. The premise of To The Moon is based around a technology that allows us to permanently reconstruct the memory on dying man. In this problem, we'll give you a chance, to implement the logic behind the scene. You‘ve been given N integers A [1], A [2],..., A [N]. On these integers, you need to implement the following operations: 1. C l r d: Adding a constant d for every {A i | l <= i <= r}, and increase the time stamp by 1, this is the only operation that will cause the time stamp increase. 2. Q l r: Querying the current sum of {A i | l <= i <= r}. 3. H l r t: Querying a history sum of {A i | l <= i <= r} in time t. 4. B t: Back to time t. And once you decide return to a past, you can never be access to a forward edition anymore. .. N, M ≤ 10 5, |A [i]| ≤ 10 9, 1 ≤ l ≤ r ≤ N, |d| ≤ 10 4 .. the system start from time 0, and the first modification is in time 1, t ≥ 0, and won't introduce you to a future state.
Input
n m A 1 A 2 ... A n ... (here following the m operations. )
Output
... (for each query, simply print the result. )
Sample Input
10 5 1 2 3 4 5 6 7 8 9 10 Q 4 4 Q 1 10 Q 2 4 C 3 6 3 Q 2 4 2 4 0 0 C 1 1 1 C 2 2 -1 Q 1 2 H 1 2 1
Sample Output
4 55 9 15 0 1
题意：
给你一个数组，让你维护，m次操作
询问当前时刻一个区间的和
询问在t时刻的一个区间的和
回到t时刻
时间+1，在此时刻+1下更新一个区间的值
题解：
函数式线段树的裸体
#include#include #include #include #include using namespace std;const int N = 5e6+10, M = 1e3+10, mod = 1000000, inf = 1e9+1000;typedef long long ll;int n,m;int l[N],r[N],root[N],add[N],tot = 0;ll sum[N];int build(int s,int t) { int now = ++tot; add[now] = 0; if(s==t) { scanf("%lld",&sum[now]); l[now] = r[now] = 0; return now; } int mid = (s+t)>>1; l[now] = build(s,mid); r[now] = build(mid+1,t); sum[now] = sum[l[now]]+sum[r[now]]; return now;}//查询在以t为根内[ll,rr]区间的值ll query(int k,int lll,int rr,int s,int t) { ll ans = (add[k]*(rr-lll+1)); if(lll==s&&rr==t) return sum[k]; int mid = (s+t)>>1; if(rr<=mid) ans+=query(l[k],lll,rr,s,mid); else if(lll>mid) ans+=query(r[k],lll,rr,mid+1,t); else { ans+=query(l[k],lll,mid,s,mid); ans+=query(r[k],mid+1,rr,mid+1,t); } return ans;}int update(int k,int lll,int rr,int d,int s,int t) { int now = ++tot; l[now] = l[k]; r[now] = r[k]; add[now] = add[k]; sum[now] = sum[k]; sum[now]+=(ll) (d*(rr-lll+1)); if(lll==s&&rr==t) { add[now]+=d; return now; } int mid = (s+t)>>1; if(rr<=mid) l[now] = update(l[k],lll,rr,d,s,mid); else if(lll>mid) r[now] = update(r[k],lll,rr,d,mid+1,t); else { l[now] = update(l[k],lll,mid,d,s,mid); r[now] = update(r[k],mid+1,rr,d,mid+1,t); } return now;}void solve() { tot = 0; root[0] = build(1,n); int now = 0; for(int i=1;i<=m;i++) { char ch[3]; scanf("%s",ch); if(ch[0]=='Q') { int a,b; scanf("%d%d",&a,&b); printf("%lld\n",query(root[now],a,b,1,n)); } else if(ch[0]=='C') { int a,b,d; scanf("%d%d%d",&a,&b,&d); root[now+1] = update(root[now],a,b,d,1,n); now++; } else if(ch[0]=='H') { int a,b,t; scanf("%d%d%d",&a,&b,&t); printf("%lld\n",query(root[t],a,b,1,n)); } else scanf("%d",&now); }}int main() { while(scanf("%d%d",&n,&m)!=EOF) { solve(); } return 0;}