标签：int val 可持久化线段树 out unique return algorithm 树状数组 register
模板太多了，写啥更新啥，顺便当自己的存档
码风很乱，见谅
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#include <iostream> #include <cstdio> #include <algorithm> #define ll long long #define rg register #define l() i&-i using namespace std; int n,m,a[500010],tree[500010],c,x,y; inline int lowbit(int x) { return x&(-x); } inline void updata(int x,int k) { for(int i=x;i<=n;i+=lowbit(x)) tree[i]+=k; } inline int query(int x) { int ans=0; for(int i=x;i;i-=lowbit(x)) ans+=tree[i]; return ans; } inline int read() { int x=0,f=1; char ch; while (ch < ‘0‘ || ch > ‘9‘) {if (ch==‘-‘)f = -1;ch = getchar();} while (ch >= ‘0‘ && ch <= ‘9‘){x = x*10+ch-‘0‘;ch = getchar();} return f*x; } int main() { n=read(); m=read(); for(int i=1;i<=n;i++) { a[i]=read(); updata(i,a[i]); } for(int i=1;i<=m;i++) { c=read(); if(c==1) { x=read(); y=read(); updata(x,y); } else { x=read(); y=read(); cout<<query(y)-query(x-1)<<endl; } } }
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#include <iostream> #include <cstdio> #include <algorithm> #define ll long long #define rg register #define l() i&-i using namespace std; int n,x,y,z,k,tree[500010],a,m,last=0; inline int read() { int x=0,f=1; char ch; while (ch < ‘0‘ || ch > ‘9‘) {if (ch==‘-‘)f = -1;ch = getchar();} while (ch >= ‘0‘ && ch <= ‘9‘){x = x*10+ch-‘0‘;ch = getchar();} return f*x; } inline int lowbit(int x) { return x&(-x); } inline void add(int x,int k) { for(int i=x;i<=n;i+=lowbit(i)) tree[i]+=k; } inline long long query(int x) { long long ans=0; for(int i=x;i;i-=lowbit(i)) ans+=tree[i]; return ans; } int main() { n=read(),m=read(); for(int i=1;i<=n;i++) { a=read(); add(i,a-last); last=a; } for(int i=1;i<=m;i++) { cin>>x; if(x==1) { cin>>y>>z>>k; add(y,k); add(z+1,-k); } if(x==2) { cin>>y; cout<<query(y)<<endl; } } }
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#include <iostream> #include <cstdio> #include <algorithm> #define ll long long #define rg register using namespace std; inline int read() { int x=0,f=1; char ch; while (ch < ‘0‘ || ch > ‘9‘) {if (ch==‘-‘)f = -1;ch = getchar();} while (ch >= ‘0‘ && ch <= ‘9‘){x = x*10+ch-‘0‘;ch = getchar();} return f*x; } int n,q,m,T[200001],tot,a[200001],b[200001]; struct cym{ int l,r,val; }tree[8000001]; inline int bulid(int l,int r)// 建树操作 { int bj=++tot; tree[bj].val=0; int mid=(l+r)>>1; if(l>r) { tree[bj].l=bulid(l,mid); tree[bj].r=bulid(mid+1,r); } return bj; } inline int change(int pre,int l,int r,int zone)// 修改 { int bj=++tot; tree[bj].l=tree[pre].l; tree[bj].r=tree[pre].r; tree[bj].val=tree[pre].val+1; int mid=(l+r)>>1; if(l<r) { if(zone<=mid) tree[bj].l=change(tree[pre].l,l,mid,zone); else tree[bj].r=change(tree[pre].r,mid+1,r,zone); } return bj; } inline int cx(int from,int to,int l,int r,int k)// 查询 { if(l==r) return l; int x=tree[tree[to].l].val-tree[tree[from].l].val; int mid=(l+r)>>1; if(x>=k) return cx(tree[from].l,tree[to].l,l,mid,k); else return cx(tree[from].r,tree[to].r,mid+1,r,k-x); } int main() { n=read(),q=read(); for(rg int i=1;i<=n;i++) { a[i]=read(); b[i]=a[i]; } sort(b+1,b+n+1);// 排序 m=unique(b+1,b+n+1)-b-1;// 看一共有多少个不重复的元素 T[0]=bulid(1,m);// 把最后一个点建在 主席树数组T[0]的位置 for(rg int i=1;i<=n;i++) { int t=lower_bound(b+1,b+m+1,a[i])-b;// 第一个比a[i] 大的树的位置 T[i]=change(T[i-1],1,m,t); } for(rg int i=1;i<=q;i++) { int l=read(),r=read(),k=read(); int anss=cx(T[l-1],T[r],1,m,k); printf("%d\n",b[anss]); } }
标签：int val 可持久化线段树 out unique return algorithm 树状数组 register
原文地址：https://www.cnblogs.com/_Yrh/p/10339620.html
