【线段树】 SPOJ 2713 Can you answer these queries IV

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更新操作：一段区间内的全部数开根号

查询操作：一段区间内的sum值

在一个值更新操作十几次之后会变成 1 之后就不需要再更新这段了

所以在update 某个区间 r-l+1==sum[rt] j就return

#include <cstdio>#include <cstring>#include <cstdlib>#include <string>#include <iostream>#include <algorithm>#include <sstream>#include <cmath>using namespace std;#include <queue>#include <stack>#include <vector>#include <deque>#define cler(arr, val)    memset(arr, val, sizeof(arr))#define FOR(i,a,b)  for(int i=a;i<=b;i++)#define IN   freopen ("in.txt" , "r" , stdin);#define OUT  freopen ("out.txt" , "w" , stdout);typedef long long  LL;const int MAXN = 120000;const int MAXM = 200000;const int INF = 0x3f3f3f3f;const int mod = 1000000007;const double eps= 1e-8;#define lson l,m, rt<<1#define rson m+1,r,rt<<1|1LL sum[MAXN<<2];void pushup(int rt){    sum[rt]=sum[rt<<1]+sum[rt<<1|1];}void build(int l,int r,int rt){    if(l==r)    {        scanf("%lld",&sum[rt]);        return ;    }    int m=(l+r)>>1;    build(lson);    build(rson);    pushup(rt);}LL query(int L,int R,int l,int r,int rt){    if(L<=l&&r<=R)        return sum[rt];    int m=(l+r)>>1;    LL res=0;    if(L<=m) res+=query(L,R,lson);    if(R>m) res+=query(L,R,rson);    return res;}void update(int L,int R,int l,int r,int rt){    if(r-l+1==sum[rt]&&L<=l && r<=R)        return ;    if(l==r)    {        sum[rt]=(LL)sqrt(sum[rt]*1.0);        return ;    }    int m=(l+r)>>1;    if(L<=m) update(L,R,lson);    if(R>m) update(L,R,rson);    pushup(rt);}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);    //freopen("out.txt", "w", stdout);#endif    int m,n,cas=0,flag=0;    while(scanf("%lld",&n)!=EOF)    {        build(1,n,1);        scanf("%d",&m);        printf("Case #%d:\n",++cas);        for(int i=0;i<m;i++)        {            int ls,rs,op;            scanf("%d%d%d",&op,&ls,&rs);            if(rs<ls)swap(ls,rs);            if(op)                printf("%lld\n",query(ls,rs,1,n,1));            else update(ls,rs,1,n,1);        }        puts("");    }    return 0;}

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