6
2 3 3 3 3 3

3
1 6
1 6
2 6

Data Constraint

分析

這題我TMD想復雜了

其實只用線段樹記錄區間最小值下標，然後分治一波就行，顯然每次ans累加上最小值min減當前區間已減量delta

比賽時寫了一波O3 inline register等卡時操作（並沒有什麽作用）

#pragma GCC optimize(3)
#include 
 
<iostream>
#include <cstdio>
#include <memory.h>
#include <queue>
#define l(x) t[x].l
#define r(x) t[x].r
#define m(x) t[x].mn
#define s(x) t[x].pos
using namespace std;
const int N=1e5+1;
struct Node {
    int l,r,mn,pos;
}t[4*N];
struct Adep {
    int l,r,dat;
};
int rt=1;
 
int a[N];
int ansl[N],ansr[N],cnt;
int n;

inline Adep In(int x,int y,int z) {
    Adep a;a.l=x;a.r=y;a.dat=z;
    return a;
}
inline void Build(int x,int l,int r) {
    l(x)=l;r(x)=r;
    if (l==r) {
        m(x)=a[l];
        s(x)=l;
        return;
    }
    int mid=l+r>>1;
    Build(x*2,l,mid);Build(x*2+1,mid+1,r);
    m(x)=m(x*2);s(x)=s(x*2);
    if (m(x)>m(x*2+1)) m(x)=m(x*2+1),s(x)=s(x*2+1);
}

inline int Query(int x,int l,int r) {
    if (l>r(x)||r<l(x)) return 0;
    if (l<=l(x)&&r(x)<=r) return s(x);
    int mid=l(x)+r(x)>>1,ans=0,val=2147483647;
    if (l<=mid) ans=Query(x*2,l,r),val=a[ans];
    if (mid<r) {
        int d=Query(x*2+1,l,r);
        if (val>a[d]) val=a[d],ans=d;
    }
    return ans;
}

inline void Solve() {
    queue <Adep> q;
    while (!q.empty()) q.pop();
    q.push(In(1,n,0));
    while (!q.empty()) {
        Adep d=q.front();q.pop();
        int l=d.l,r=d.r,dt=d.dat;
        int o=Query(rt,l,r),m=a[o],p=0;
        m-=dt;
        while (m) {
            ansl[++cnt]=l;ansr[cnt]=r;
            m--;p++;
        }
        if (l<=o-1) q.push(In(l,o-1,dt+p));
        if (o+1<=r) q.push(In(o+1,r,dt+p));
    }
}

int main() {
    freopen("range.in","r",stdin);
    freopen("range.out","w",stdout);
    scanf("%d",&n);
    for (register int i=1;i<=n;i++) scanf("%d",&a[i]);
    for (register int i=1;i<=4*n;i++) t[i].mn=2147483647;
    a[0]=2147483647;
    Build(rt,1,n);
    Solve();
    printf("%d\n",cnt);
    for (register int i=1;i<=cnt;i++) printf("%d %d\n",ansl[i],ansr[i]);
    fclose(stdin);fclose(stdout);
}

[線段樹]JZOJ 5812